Algebra Quiz: Zeros Of Polynomials To Construct Graphs

What this quiz covers

This quiz focuses on Zeros Of Polynomials To Construct Graphs, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Zeros: $x=-2,-1,1,2$; x-intercepts: $(-2,0),(-1,0),(1,0),(2,0)$; end behavior: both ends up; crosses at each zero. (correct answer)
2. Zeros: $x=-2,-1,1,2$; x-intercepts: $(-2,0),(-1,0),(1,0),(2,0)$; end behavior: left down, right up; crosses at each zero.
3. Zeros: $x=-2,1,2$; x-intercepts: $(-2,0),(1,0),(2,0)$; end behavior: both ends up.
4. Zeros: $x=-2,-1,1,2$; x-intercepts: $(-2,0),(-1,0),(1,0),(2,0)$; end behavior: both ends down; crosses at each zero.

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. To construct a rough sketch from zeros: (1) Mark the zeros on the x-axis, (2) Determine end behavior from degree and leading coefficient sign, (3) Connect the zeros with a smooth curve that crosses/touches appropriately and has the right end behavior. You don't need exact heights—just show the general shape, where it crosses the x-axis, and which direction the ends go! To sketch P(x) = (x - 2)(x + 2)(x - 1)(x + 1): (1) Zeros are at x = 2, -2, 1, -1, so mark these on the x-axis. (2) This polynomial has degree 4 (count the factors or highest power) with leading coefficient positive, so end behavior is both ends up. (3) At each zero, check multiplicity: all are multiplicity 1 (odd), so crosses at each. (4) Connect with smooth curve showing up to 3 turns, starting and ending with correct end behavior. The sketch doesn't need exact heights, just the right shape! Choice A correctly shows sketch with proper crossings and end behavior by using degree 4 (even) and positive leading coefficient for both ends up, identifying all zeros. Choice C has the end behavior backwards: with degree 4 (even) and leading coefficient positive, the ends should both go up, not left down and right up. Remember even degree means both ends same direction. The sign of the leading coefficient then determines up or down! Quick check: count your zeros (including multiplicities) and it should equal the degree. If P(x) is degree 4, you should find 4 zeros total (could be 4 simple zeros, or 1 with multiplicity 2 and 2 simple, etc.). If your count doesn't match the degree, you've either missed a zero or the polynomial isn't completely factored. This check prevents forgetting zeros!

Question 2

1. Zeros at $x=-1,2,4$; as $x\to -\infty$, $P(x)\to -\infty$ and as $x\to \infty$, $P(x)\to \infty$
2. Zeros at $x=1,-2,-4$; as $x\to -\infty$, $P(x)\to \infty$ and as $x\to \infty$, $P(x)\to -\infty$
3. Zeros at $x=-1,2,4$; as $x\to -\infty$, $P(x)\to \infty$ and as $x\to \infty$, $P(x)\to -\infty$ (correct answer)
4. Zeros at $x=-1,2,4$; as $x\to -\infty$, $P(x)\to \infty$ and as $x\to \infty$, $P(x)\to \infty$

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. A polynomial's end behavior (what happens as x → ∞ and x → -∞) is determined by its degree and leading coefficient: for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree polynomials, the ends go opposite directions (if positive leading coefficient: left end down, right end up). This end behavior, combined with zeros, gives you the rough shape! From P(x) = -(x+1)(x-2)(x-4), we find zeros by setting each factor equal to zero: (x+1) = 0 → x = -1, (x-2) = 0 → x = 2, (x-4) = 0 → x = 4. The zeros are x = -1, 2, 4. This polynomial has degree 3 (three factors multiplied) with leading coefficient -1 (negative from the minus sign out front). Since degree 3 is odd and the leading coefficient is negative, the end behavior is: as x → -∞ (far left), P(x) → ∞, and as x → ∞ (far right), P(x) → -∞. Choice C correctly identifies zeros as -1, 2, 4 and shows end behavior with left end going to ∞ and right end going to -∞, properly using the negative leading coefficient and odd degree. Choice A has the end behavior backwards: with degree 3 (odd) and leading coefficient negative, the ends should go opposite directions with left up and right down, not left down and right up. Remember odd degree means opposite directions. The sign of the leading coefficient then determines up or down! End behavior memory tricks: Even degree polynomials make 'U-shapes' or 'n-shapes' (both ends same direction), while odd degree polynomials make 'chair shapes' or 'S-curves' (ends opposite). Positive leading coefficient: right end goes up. Negative: right end goes down. Combine these: degree 3 with negative leading coefficient = left up, right down (like sitting in an upside-down chair). Visual mnemonics help!

Question 3

1. Zeros: $x=0$ (mult. 2), $x=2$ (mult. 1), $x=-1$ (mult. 1); touches at $x=0$, crosses at $x=2$ and $x=-1$
2. Zeros: $x=0$ (mult. 1), $x=2$ (mult. 2), $x=-1$ (mult. 1); crosses at $x=0$ and $x=-1$, touches at $x=2$ (correct answer)
3. Zeros: $x=0$ (mult. 1), $x=2$ (mult. 2), $x=-1$ (mult. 1); touches at $x=0$ and $x=2$, crosses at $x=-1$
4. Zeros: $x=0$ (mult. 1), $x=2$ (mult. 1), $x=-1$ (mult. 2); crosses at $x=0$ and $x=2$, touches at $x=-1$

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. The multiplicity of a zero (how many times a factor appears) tells you how the graph behaves there: if a zero has odd multiplicity (like just (x - 2) or (x - 2)³), the graph crosses the x-axis at that point. If a zero has even multiplicity (like (x - 2)² or (x - 2)⁴), the graph touches the x-axis but bounces back without crossing—it turns around at that zero! The polynomial P(x) = x(x-2)^2(x+1) has a zero at x=2 with multiplicity 2 because the factor (x-2) appears 2 times. Since 2 is even, the graph touches and turns at this zero. This is different from simple zeros where the graph just crosses straight through. Higher multiplicity means the graph 'hugs' the x-axis more at that zero! Choice A correctly identifies zeros as x=0 (mult. 1), x=2 (mult. 2), x=-1 (mult. 1) and describes behavior as crosses at x=0 and x=-1, touches at x=2 by recognizing multiplicity effects. Choice B has the multiplicity wrong: x=0 has multiplicity 1 (odd, crosses), not 2 (even, touches). Check the power on each factor! Multiplicity matters: Simple zero (appears once) = graph crosses straight through. Even multiplicity (appears 2, 4, 6... times) = graph touches and bounces without crossing, creating a turning point at that zero. Odd multiplicity higher than 1 (appears 3, 5... times) = graph crosses but flattens out at that zero. The more times a factor repeats, the 'flatter' the graph gets at that zero!

Question 4

1. x-intercepts: $(-1,0),(3,0),(4,0)$; end behavior: left down, right up
2. x-intercepts: $(1,0),(-3,0),(4,0)$; end behavior: left down, right up (correct answer)
3. x-intercepts: $(1,0),(-3,0),(4,0)$; end behavior: left up, right down
4. x-intercepts: $(1,0),(-3,0)$; end behavior: left down, right up

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂)(x - r₃), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂, r₃. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x-1)(x+3)(x-4), we find zeros by setting each factor equal to zero: (x-1)=0 → x=1, (x+3)=0 → x=-3, (x-4)=0 → x=4. The zeros are x=1, -3, 4. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 3) gives zero at x = 3, and (x + 5) = (x - (-5)) gives zero at x = -5. Choice B correctly identifies zeros as x=1, -3, 4 (x-intercepts (1,0), (-3,0), (4,0)) and end behavior left down, right up by properly applying zero product property and using degree and leading coefficient. Choice A has the sign wrong on zeros: from the factor (x+3) = (x - (-3)), the zero is x = -3, not x = 3. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

Question 5

1. Zeros at $x=2$ (mult. 2), $x=-1$, $x=-3$; crosses at $x=2$, touches at $x=-1$ and $x=-3$; both ends up
2. Zeros at $x=2$ (mult. 2), $x=-1$, $x=-3$; touches at $x=2$, crosses at $x=-1$ and $x=-3$; both ends up (correct answer)
3. Zeros at $x=2$ (mult. 2), $x=-1$, $x=-3$; touches at $x=2$; both ends down
4. Zeros at $x=2$ (mult. 2), $x=1$, $x=3$; touches at $x=2$; both ends up

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. The multiplicity of a zero (how many times a factor appears) tells you how the graph behaves there: if a zero has odd multiplicity (like just (x - 2) or (x - 2)³), the graph crosses the x-axis at that point. If a zero has even multiplicity (like (x - 2)² or (x - 2)⁴), the graph touches the x-axis but bounces back without crossing—it turns around at that zero! The polynomial P(x) = (x - 2)^2 (x + 1)(x + 3) has a zero at x = 2 with multiplicity 2 because the factor (x - 2) appears 2 times. Since 2 is even, the graph touches and turns at this zero. This is different from simple zeros where the graph just crosses straight through. Higher multiplicity means the graph 'hugs' the x-axis more at that zero! Zeros at x = -1 and x = -3 have multiplicity 1 (odd), so crosses there. Degree 4 even positive, both ends up. Choice A correctly describes behavior as touches at x=2, crosses at x=-1 and x=-3 with both ends up by recognizing multiplicity effects and using degree and leading coefficient. Choice D has the end behavior backwards: with degree 4 (even) and leading coefficient positive, the ends should both up, not both down. Remember even degree means both ends same direction. The sign of the leading coefficient then determines up or down! Multiplicity matters: Simple zero (appears once) = graph crosses straight through. Even multiplicity (appears 2, 4, 6... times) = graph touches and bounces without crossing, creating a turning point at that zero. Odd multiplicity higher than 1 (appears 3, 5... times) = graph crosses but flattens out at that zero. The more times a factor repeats, the 'flatter' the graph gets at that zero! Quick check: count your zeros (including multiplicities) and it should equal the degree. If P(x) is degree 4, you should find 4 zeros total (could be 4 simple zeros, or 1 with multiplicity 2 and 2 simple, etc.). If your count doesn't match the degree, you've either missed a zero or the polynomial isn't completely factored. This check prevents forgetting zeros!

Question 6

1. Zeros: $x=4,2,1$; x-intercepts: $(4,0),(2,0),(1,0)$; end behavior: left up, right down; crosses at each zero.
2. Zeros: $x=4,-2,1$; x-intercepts: $(4,0),(-2,0),(1,0)$; end behavior: left down, right up; crosses at each zero.
3. Zeros: $x=4,-2,1$; x-intercepts: $(4,0),(-2,0),(1,0)$; end behavior: both ends down; crosses at each zero.
4. Zeros: $x=4,-2,1$; x-intercepts: $(4,0),(-2,0),(1,0)$; end behavior: left up, right down; crosses at each zero. (correct answer)

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. A polynomial's end behavior (what happens as $x \to \infty$ and $x \to -\infty$) is determined by its degree and leading coefficient: for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree polynomials, the ends go opposite directions (if positive leading coefficient: left end down, right end up). This end behavior, combined with zeros, gives you the rough shape! This polynomial has degree 3 and leading coefficient $-1$. Since 3 is odd and $-1$ is negative, the end behavior is: as $x \to -\infty$ (far left), P(x) $\to \infty$, and as $x \to \infty$ (far right), P(x) $\to -\infty$. Think of it this way: odd degree with negative leading coefficient means left up, right down. This end behavior plus the zeros gives us the skeleton of the graph! Choice A correctly identifies zeros as $x=4, -2, 1$ and shows sketch with proper crossings and end behavior by using degree and leading coefficient. Choice C has the end behavior backwards: with degree 3 (odd) and leading coefficient negative, the ends should be left up and right down, not left down and right up. Remember odd degree means opposite directions. The sign of the leading coefficient then determines up or down! End behavior memory tricks: Even degree polynomials make 'U-shapes' or 'n-shapes' (both ends same direction), while odd degree polynomials make 'chair shapes' or 'S-curves' (ends opposite). Positive leading coefficient: right end goes up. Negative: right end goes down. Combine these: degree 3 with positive leading coefficient = left down, right up (like sitting in a chair). Degree 4 with negative leading coefficient = both ends down (like an upside-down U). Visual mnemonics help!

Question 7

1. Break-even x-values: $-2,1,5$; x-intercepts: $(-2,0),(1,0),(5,0)$ (correct answer)
2. Break-even x-values: $2,-1,-5$; x-intercepts: $(2,0),(-1,0),(-5,0)$
3. Break-even x-values: $-2,1$; x-intercepts: $(-2,0),(1,0)$
4. Break-even x-values: $-2,5$; x-intercepts: $(-2,0),(5,0)$

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂)(x - r₃), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂, r₃. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x-1)(x-5)(x+2), we find zeros by setting each factor equal to zero: (x-1) = 0 → x = 1, (x-5) = 0 → x = 5, (x+2) = 0 → x = -2. The zeros (break-even points) are x = -2, 1, 5. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 3) gives zero at x = 3, and (x + 5) = (x - (-5)) gives zero at x = -5. Choice A correctly identifies break-even x-values as -2, 1, 5 and shows x-intercepts as (-2,0), (1,0), (5,0) by properly applying the zero product property to find where profit equals zero. Choice B has the signs wrong on all zeros: from the factor (x-1), the zero is x = 1, not x = -1. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

Question 8

1. x-intercepts at $(-2,0)$, $(1,0)$, $(3,0)$; end behavior: left up, right down (correct answer)
2. x-intercepts at $(-2,0)$ and $(3,0)$ only; end behavior: left up, right down
3. x-intercepts at $(2,0)$, $(1,0)$, $(3,0)$; end behavior: left up, right down
4. x-intercepts at $(-2,0)$, $(1,0)$, $(3,0)$; end behavior: left down, right up

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. A polynomial's end behavior (what happens as x → ∞ and x → -∞) is determined by its degree and leading coefficient: for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree polynomials, the ends go opposite directions (if positive leading coefficient: left end down, right end up). This end behavior, combined with zeros, gives you the rough shape! This polynomial has degree 3 and leading coefficient -2. Since 3 is odd and -2 is negative, the end behavior is: as x → -∞ (far left), P(x) → ∞, and as x → ∞ (far right), P(x) → -∞. Think of it this way: odd degree with negative a means left up, right down. This end behavior plus the zeros gives us the skeleton of the graph! Zeros are at x = -2, 1, 3 from setting each factor to zero. Choice B correctly shows sketch with proper crossings and end behavior by using degree and leading coefficient. Choice A has the end behavior backwards: with degree 3 (odd) and leading coefficient negative, the ends should left up right down, not left down right up. Remember odd degree means opposite directions. The sign of the leading coefficient then determines up or down! End behavior memory tricks: Even degree polynomials make 'U-shapes' or 'n-shapes' (both ends same direction), while odd degree polynomials make 'chair shapes' or ' S-curves' (ends opposite). Positive leading coefficient: right end goes up. Negative: right end goes down. Combine these: degree 3 with positive leading coefficient = left down, right up (like sitting in a chair). Degree 4 with negative leading coefficient = both ends down (like an upside-down U). Visual mnemonics help!

Question 9

1. Zeros: $x=2,3$; x-intercepts: $(2,0)$ and $(3,0)$; opens up. (correct answer)
2. Zeros: $x=-2,-3$; x-intercepts: $(-2,0)$ and $(-3,0)$; opens up.
3. Zeros: $x=2,3$; x-intercepts: $(2,0)$ and $(3,0)$; opens down.
4. Zeros: $x=1,6$; x-intercepts: $(1,0)$ and $(6,0)$; opens up.

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x - 2)(x - 3), we find zeros by setting each factor equal to zero: (x - 2) = 0 → x = 2, (x - 3) = 0 → x = 3. The zeros are x = 2, 3. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 2) gives zero at x = 2, but if it were (x + 2), it would give x = -2. Choice A correctly identifies zeros as x=2, 3 and shows the parabola opens up by using positive leading coefficient for both ends up. Choice B has the signs wrong on the zeros: from the factors (x - 2) and (x - 3), the zeros are x = 2 and 3, not x = -2 and -3. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

Question 10

1. Zeros: $x=-3,1,3$ with $x=1$ multiplicity 2; touches at $x=1$; left down, right up
2. Zeros: $x=-3,1,3$ with $x=1$ multiplicity 2; crosses at all three zeros; both ends up
3. Zeros: $x=-3,-1,3$ with $x=-1$ multiplicity 2; touches at $x=-1$; both ends up
4. Zeros: $x=-3,1,3$ with $x=1$ multiplicity 2; touches at $x=1$ and crosses at $x=-3$ and $x=3$; both ends up (correct answer)

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. The multiplicity of a zero (how many times a factor appears) tells you how the graph behaves there: if a zero has odd multiplicity (like just $(x - 2)$ or $(x - 2)^3$), the graph crosses the x-axis at that point. If a zero has even multiplicity (like $(x - 2)^2$ or $(x - 2)^4$), the graph touches the x-axis but bounces back without crossing—it turns around at that zero! The polynomial $P(x) = (x - 3)(x + 3)(x - 1)^2$ has a zero at $x = 1$ with multiplicity 2 because the factor $(x - 1)$ appears 2 times. Since 2 is even, the graph touches and turns at this zero. This is different from simple zeros where the graph just crosses straight through. Higher multiplicity means the graph 'hugs' the x-axis more at that zero! Zeros at $x = -3$ and 3 have multiplicity 1 (odd), so crosses there. Degree 4 even positive, both ends up. Choice A correctly describes behavior as touches at $x=1$, crosses at $x=-3$ and $x=3$ with both ends up by recognizing multiplicity effects and using degree and leading coefficient. Choice C has the end behavior backwards: with degree 4 (even) and leading coefficient positive, the ends should both up, not left down right up. Remember even degree means both ends same direction. The sign of the leading coefficient then determines up or down! Multiplicity matters: Simple zero (appears once) = graph crosses straight through. Even multiplicity (appears 2, 4, 6... times) = graph touches and bounces without crossing, creating a turning point at that zero. Odd multiplicity higher than 1 (appears 3, 5... times) = graph crosses but flattens out at that zero. The more times a factor repeats, the 'flatter' the graph gets at that zero! Quick check: count your zeros (including multiplicities) and it should equal the degree. If $P(x)$ is degree 4, you should find 4 zeros total (could be 4 simple zeros, or 1 with multiplicity 2 and 2 simple, etc.). If your count doesn't match the degree, you've either missed a zero or the polynomial isn't completely factored. This check prevents forgetting zeros!

Question 11

1. Zeros: $x=4,-1,2$; x-intercepts: $(4,0),(-1,0),(2,0)$ (correct answer)
2. Zeros: $x=-4,1,-2$; x-intercepts: $(-4,0),(1,0),(-2,0)$
3. Zeros: $x=4,1,2$; x-intercepts: $(4,0),(1,0),(2,0)$
4. Zeros: $x=4,-1$; x-intercepts: $(4,0),(-1,0)$

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂)(x - r₃), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂, r₃. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x-4)(x+1)(x-2), we find zeros by setting each factor equal to zero: (x-4) = 0 → x = 4, (x+1) = 0 → x = -1, (x-2) = 0 → x = 2. The zeros are x = 4, -1, 2. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 3) gives zero at x = 3, and (x + 5) = (x - (-5)) gives zero at x = -5. Choice A correctly identifies zeros as x = 4, -1, 2 and shows x-intercepts as (4,0), (-1,0), (2,0) by properly applying the zero product property to each factor. Choice B has the signs wrong on all zeros: from the factor (x-4), the zero is x = 4, not x = -4. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

Question 12

1. Zeros: $x=3,-1,2$; x-intercepts: $(3,0), (-1,0), (2,0)$; end behavior: left down, right up; crosses at each zero. (correct answer)
2. Zeros: $x=-3,1,-2$; x-intercepts: $(-3,0), (1,0), (-2,0)$; end behavior: left down, right up; crosses at each zero.
3. Zeros: $x=3,-1$ only; x-intercepts: $(3,0), (-1,0)$; end behavior: both ends up.
4. Zeros: $x=3,-1,2$; x-intercepts: $(3,0), (-1,0), (2,0)$; end behavior: both ends up; crosses at each zero.

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like $P(x) = a(x - r_1)(x - r_2)(x - r_3)$, the zeros are immediately visible: set each factor equal to zero to get $x = r_1, r_2, r_3$. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form $P(x) = (x - 3)(x + 1)(x - 2)$, we find zeros by setting each factor equal to zero: $(x - 3) = 0 \to x = 3$, $(x + 1) = 0 \to x = -1$, $(x - 2) = 0 \to x = 2$. The zeros are $x = 3, -1, 2$. Remember: from $(x - r)$, the zero is $x = r$ (opposite sign!), so $(x - 3)$ gives zero at $x = 3$, and $(x + 1) = (x - (-1))$ gives zero at $x = -1$. Choice A correctly identifies zeros as $x=3, -1, 2$ and describes the sketch with proper crossings and end behavior by properly applying the zero product property, using degree 3 (odd) and positive leading coefficient for left down, right up. Choice B has the signs wrong on the zeros: for example, from the factor $(x + 1) = (x - (-1))$, the zero is $x = -1$, not $x = 1$. This sign flip is super common! Remember: $(x - r)$ gives zero at $x = r$, so you reverse the sign from what's in the factor. Think: what value makes $(x [sign] [number])$ equal to zero? The zero-finding procedure from factored form: for each factor $(x - r)$, set it equal to zero and solve: $(x - r) = 0 \to x = r$. That $r$ is your zero. Do this for every factor. Watch signs carefully: $(x - 3)$ gives $x = 3$, $(x + 5)$ gives $x = -5$. If a factor appears multiple times like $(x - 2)^3$, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

Question 13

1. Zeros: $x=2$ and $x=-2$; x-intercepts: $(2,0)$ and $(-2,0)$ (correct answer)
2. Zeros: $x=0$ and $x=4$; x-intercepts: $(0,0)$ and $(4,0)$
3. Zeros: $x=4$ and $x=-4$; x-intercepts: $(4,0)$ and $(-4,0)$
4. Zeros: $x=2$ only (mult. 2); x-intercept: $(2,0)$

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂)(x - r₃), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂, r₃. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x-2)(x+2), we find zeros by setting each factor equal to zero: (x-2)=0 → x=2, (x+2)=0 → x=-2. The zeros are x=2, -2. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 3) gives zero at x = 3, and (x + 5) = (x - (-5)) gives zero at x = -5. Choice B correctly identifies zeros as x=2, -2 (x-intercepts (2,0), (-2,0)) by properly applying zero product property. Choice A has zeros at the wrong x-values: 4 and -4 would come from (x-4)(x+4) = x^2 -16, not x^2 -4. When reading from factored form, carefully solve each (x - r) = 0—don't rush and assume the signs! Write out each step: (x [sign] [value]) = 0 → x = [zero value]. The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

Question 14

1. At $x=-1$ the graph crosses the x-axis and flattens; at $x=2$ the graph crosses the x-axis (correct answer)
2. At $x=-1$ the graph touches (bounces) off the x-axis; at $x=2$ the graph touches
3. At $x=-1$ the graph crosses normally (no flattening); at $x=2$ the graph touches
4. At $x=1$ the graph crosses and flattens; at $x=-2$ the graph crosses

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. The multiplicity of a zero (how many times a factor appears) tells you how the graph behaves there: if a zero has odd multiplicity (like just $(x - 2)$ or $(x - 2)^3$), the graph crosses the x-axis at that point. If a zero has even multiplicity (like $(x - 2)^2$ or $(x - 2)^4$), the graph touches the x-axis but bounces back without crossing—it turns around at that zero! The polynomial $P(x) = (x + 1)^3 (x - 2)$ has a zero at $x = -1$ with multiplicity 3 because the factor $(x + 1)$ appears 3 times. Since 3 is odd, the graph crosses but flattens at this zero. This is different from simple zeros where the graph just crosses straight through. Higher multiplicity means the graph 'hugs' the x-axis more at that zero! Zero at $x = 2$ has multiplicity 1 (odd), so crosses normally. Choice A correctly describes behavior as crosses and flattens at $x = -1$, crosses at $x = 2$ by recognizing multiplicity effects. Choice B shows the graph touching at $x = -1$, but this zero has odd multiplicity 3, so the graph should cross through (with flattening), not touch and bounce. Even multiplicities create turnarounds at zeros, while odd multiplicities allow crossings. Check the power on each factor! Multiplicity matters: Simple zero (appears once) = graph crosses straight through. Even multiplicity (appears 2, 4, 6... times) = graph touches and bounces without crossing, creating a turning point at that zero. Odd multiplicity higher than 1 (appears 3, 5... times) = graph crosses but flattens out at that zero. The more times a factor repeats, the 'flatter' the graph gets at that zero!

Question 15

1. The graph has x-intercepts at $x = 0, 1, 3$ and crosses the x-axis at all three points, with the y-intercept at $(0, 0)$ (correct answer)
2. The graph has x-intercepts at $x = 0, 1, 3$ but touches without crossing at $x = 0$ due to the factored form having $x^2$
3. The graph has x-intercepts at $x = 1, 3$ only, since the factor of $x$ doesn't create a visible intercept on the graph
4. The graph has x-intercepts at $x = 0, 1, 3$ and passes through the origin since one factor is $x$

Explanation: Factoring: $s(x) = 2x^3 - 8x^2 + 6x = 2x(x^2 - 4x + 3) = 2x(x-1)(x-3)$. The zeros are at $x = 0, 1, 3$, each appearing once in the factored form, so the graph crosses the x-axis at all three points. The y-intercept is at $(0,0)$ since $s(0) = 0$. Choice B incorrectly suggests touching behavior at $x = 0$. Choice C incorrectly excludes $x = 0$ as an intercept. Choice D is partially correct but doesn't fully describe the crossing behavior.

Question 16

1. Two x-intercepts, because $x^2 - 1$ and $x^2 - 4$ each contribute one zero to the polynomial function
2. Three x-intercepts, because the polynomial can be written in the form $(x - a)(x - b)(x - c)$ for some values
3. Four x-intercepts, because both $x^2 - 1 = (x-1)(x+1)$ and $x^2 - 4 = (x-2)(x+2)$ factor further (correct answer)
4. Five x-intercepts, because the original polynomial $x^4 - 5x^2 + 4$ is degree 4 and has an additional repeated root

Explanation: The polynomial $p(x) = (x^2-1)(x^2-4)$ can be factored completely as $p(x) = (x-1)(x+1)(x-2)(x+2)$. Each linear factor corresponds to one x-intercept, giving four x-intercepts at $x = -2, -1, 1, 2$. Choice A incorrectly assumes each quadratic factor gives one zero. Choice B gives an incorrect count. Choice D incorrectly suggests five intercepts and mentions a repeated root that doesn't exist.

Question 17

1. Zeros: $x=\pm1,\pm2$; x-intercepts: $(-2,0),(-1,0),(1,0),(2,0)$; end behavior: left down, right up.
2. Zeros: $x=\pm1,\pm2$; x-intercepts: $(-2,0),(-1,0),(1,0),(2,0)$; end behavior: both ends up; crosses at each zero. (correct answer)
3. Zeros: $x=\pm1,\pm4$; x-intercepts: $(-4,0),(-1,0),(1,0),(4,0)$; end behavior: both ends up.
4. Zeros: $x=\pm1,\pm2$; x-intercepts: $(-2,0),(-1,0),(1,0),(2,0)$; end behavior: both ends down.

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂)(x - r₃)(x - r₄), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂, r₃, r₄. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x - 2)(x + 2)(x - 1)(x + 1), we find zeros by setting each factor equal to zero: (x - 2) = 0 → x = 2, (x + 2) = 0 → x = -2, (x - 1) = 0 → x = 1, (x + 1) = 0 → x = -1. The zeros are x = ±2, ±1. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 2) gives zero at x = 2, and (x + 2) = (x - (-2)) gives zero at x = -2. Choice A correctly identifies zeros as x=±1, ±2 and shows sketch with proper crossings and end behavior by properly applying zero product property, using degree 4 (even) and positive leading for both ends up. Choice B has zeros at the wrong x-values: it shows ±1, ±4, but the factors give ±1, ±2. When reading from factored form, carefully solve each (x - r) = 0—don't rush and assume the signs! Write out each step: (x [sign] [value]) = 0 → x = [zero value]. The four-step polynomial sketching recipe: (1) Find and mark all zeros on the x-axis, (2) Determine end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive = eventually up, negative = eventually down), (3) For each zero, check if it crosses (odd multiplicity) or touches (even multiplicity), (4) Connect with smooth curve that goes through/touches all zeros with correct end behavior. Don't worry about exact curve—just the general shape!

Question 18

1. Zeros: $x=0$ (mult. 1), $x=2$ (mult. 2), $x=-1$ (mult. 1); crosses at $x=0$ and $x=-1$, touches at $x=2$; end behavior: both ends up. (correct answer)
2. Zeros: $x=0$ (mult. 2), $x=2$ (mult. 1), $x=-1$ (mult. 1); touches at $x=0$, crosses at $x=2$ and $x=-1$; end behavior: both ends up.
3. Zeros: $x=0$ (mult. 1), $x=2$ (mult. 2), $x=-1$ (mult. 1); touches at all zeros; end behavior: both ends up.
4. Zeros: $x=0$ (mult. 1), $x=2$ (mult. 2), $x=-1$ (mult. 1); crosses at $x=0$ and $x=-1$, touches at $x=2$; end behavior: left down, right up.

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. The multiplicity of a zero (how many times a factor appears) tells you how the graph behaves there: if a zero has odd multiplicity (like just (x - 2) or (x - 2)³), the graph crosses the x-axis at that point. If a zero has even multiplicity (like (x - 2)² or (x - 2)⁴), the graph touches the x-axis but bounces back without crossing—it turns around at that zero! The polynomial P(x) = x(x - 2)²(x + 1) has a zero at x = 2 with multiplicity 2 because the factor (x - 2) appears 2 times. Since 2 is even, the graph touches and turns at this zero. This is different from simple zeros where the graph just crosses straight through. Higher multiplicity means the graph 'hugs' the x-axis more at that zero! Choice A correctly describes behavior as crosses at x=0 and x=-1, touches at x=2 by recognizing multiplicity effects, with degree 4 (even) and positive leading for both ends up. Choice D has the end behavior backwards: with degree 4 (even) and leading coefficient positive, the ends should both go up, not left down and right up. Remember even degree means both ends same direction. The sign of the leading coefficient then determines up or down! Multiplicity matters: Simple zero (appears once) = graph crosses straight through. Even multiplicity (appears 2, 4, 6... times) = graph touches and bounces without crossing, creating a turning point at that zero. Odd multiplicity higher than 1 (appears 3, 5... times) = graph crosses but flattens out at that zero. The more times a factor repeats, the 'flatter' the graph gets at that zero!

Question 19

1. Break-even x-values: $x=-2,1,5$; x-intercepts: $(-2,0),(1,0),(5,0)$; end behavior: both ends up.
2. Break-even x-values: $x=-2,1$ only; x-intercepts: $(-2,0),(1,0)$; end behavior: left down, right up.
3. Break-even x-values: $x=-2,1,5$; x-intercepts: $(-2,0),(1,0),(5,0)$; end behavior: left down, right up; crosses at each intercept. (correct answer)
4. Break-even x-values: $x=2,-1,-5$; x-intercepts: $(2,0),(-1,0),(-5,0)$; end behavior: left down, right up.

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like $P(x) = a(x - r_1)(x - r_2)(x - r_3)$, the zeros are immediately visible: set each factor equal to zero to get $x = r_1, r_2, r_3$. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form $P(x) = (x - 5)(x - 1)(x + 2)$, we find zeros by setting each factor equal to zero: $(x - 5) = 0 \rightarrow x = 5$, $(x - 1) = 0 \rightarrow x = 1$, $(x + 2) = 0 \rightarrow x = -2$. The zeros are $x = 5, 1, -2$. Remember: from $(x - r)$, the zero is $x = r$ (opposite sign!), so $(x - 5)$ gives zero at $x = 5$, and $(x + 2) = (x - (-2))$ gives zero at $x = -2$. Choice A correctly identifies break-even as $x=-2, 1, 5$ and shows sketch with proper crossings and end behavior by properly applying zero product property, using degree 3 (odd) and positive leading for left down, right up. Choice B has the signs wrong on the zeros: for example, from the factor $(x + 2) = (x - (-2))$, the zero is $x = -2$, not $x = 2$. This sign flip is super common! Remember: $(x - r)$ gives zero at $x = r$, so you reverse the sign from what's in the factor. Think: what value makes $(x \text{ [sign]} \text{ [number]})$ equal to zero? The zero-finding procedure from factored form: for each factor $(x - r)$, set it equal to zero and solve: $(x - r) = 0 \rightarrow x = r$. That r is your zero. Do this for every factor. Watch signs carefully: $(x - 3)$ gives $x = 3$, $(x + 5)$ gives $x = -5$. If a factor appears multiple times like $(x - 2)^3$, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

Question 20

1. x-intercepts: $(-3,0)$ and $(1,0)$; crosses at $x=-3$ and touches at $x=1$; end behavior: left down, right up.
2. x-intercepts: $(-3,0)$ only; touches at $x=-3$; end behavior: both ends up.
3. x-intercepts: $(-3,0)$ and $(1,0)$; touches at $x=-3$ and crosses at $x=1$; end behavior: left down, right up. (correct answer)
4. x-intercepts: $(3,0)$ and $(1,0)$; touches at $x=3$ and crosses at $x=1$; end behavior: left down, right up.

Explanation: This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. The multiplicity of a zero (how many times a factor appears) tells you how the graph behaves there: if a zero has odd multiplicity (like just (x - 2) or (x - 2)³), the graph crosses the x-axis at that point. If a zero has even multiplicity (like (x - 2)² or (x - 2)⁴), the graph touches the x-axis but bounces back without crossing—it turns around at that zero! The polynomial P(x) = (x + 3)²(x - 1) has a zero at x = -3 with multiplicity 2 because the factor (x + 3) appears 2 times. Since 2 is even, the graph touches and turns at this zero. This is different from simple zeros where the graph just crosses straight through. Higher multiplicity means the graph 'hugs' the x-axis more at that zero! Choice A correctly describes behavior as touches at x=-3 and crosses at x=1 by recognizing multiplicity effects, with degree 3 (odd) and positive leading for left down, right up. Choice D has the sign wrong on a zero: from the factor (x + 3) = (x - (-3)), the zero is x = -3, not x = 3. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? Multiplicity matters: Simple zero (appears once) = graph crosses straight through. Even multiplicity (appears 2, 4, 6... times) = graph touches and bounces without crossing, creating a turning point at that zero. Odd multiplicity higher than 1 (appears 3, 5... times) = graph crosses but flattens out at that zero. The more times a factor repeats, the 'flatter' the graph gets at that zero!