Zeros of Polynomials to Construct Graphs - Algebra 1

Consider the quartic polynomial $P(x)=(x-2)(x+2)(x-1)(x+1).$ Identify all zeros and use them to sketch a rough graph, including correct end behavior.
Answer: 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. 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!
Factor $P(x)=x^2-5x+6$ and use the zeros to sketch a rough parabola. Your sketch should show the x-intercepts and whether the parabola opens up or down.
Answer: Zeros: $x=2,3$ ; x-intercepts: $(2,0)$ and $(3,0)$ ; opens up. 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!
Use the factored form $P(x)=-(x-4)(x+2)(x-1)$ to identify the zeros and sketch a rough graph, showing x-intercepts and correct end behavior.
Answer: Zeros: $x=4,-2,1$ ; x-intercepts: $(4,0),(-2,0),(1,0)$ ; end behavior: left up, right down; crosses at each zero. 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!
A polynomial is given by $P(x)=(x-3)^2(x+2)$ . What are the x-intercepts, and how does the graph behave at each intercept (crosses or touches)?
Answer: x-intercepts: $(3,0)$ and $(-2,0)$ ; touches at $x=3$ and crosses at $x=-2$ 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! From the factored form P(x) = (x-3)²(x+2), we find zeros by setting each factor equal to zero: (x-3)² = 0 → x = 3 (multiplicity 2), (x+2) = 0 → x = -2 (multiplicity 1). The x-intercepts are at (3,0) and (-2,0). Since the zero at x = 3 has even multiplicity (2), the graph touches and turns at this point. Since the zero at x = -2 has odd multiplicity (1), the graph crosses through at this point. Choice B correctly identifies x-intercepts at (3,0) and (-2,0) and states that the graph touches at x = 3 and crosses at x = -2 by recognizing that even multiplicity creates a touch point and odd multiplicity creates a crossing. Choice A incorrectly states the graph crosses at both zeros, but at x = 3 with multiplicity 2 (even), the graph must touch and bounce, not cross through. 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!
Factor $P(x)=x^2-4$ and use the zeros to sketch a rough graph. Which option correctly identifies the zeros and x-intercepts?
Answer: Zeros: $x=2$ and $x=-2$ ; x-intercepts: $(2,0)$ and $(-2,0)$ 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!
Factor $P(x)=x^2+3x-10$ and use the zeros to determine the x-intercepts of the graph of $y=P(x)$ .
Answer: x-intercepts: $(-5,0),(2,0)$ 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. To factor P(x) = x² + 3x - 10, we need two numbers that multiply to -10 and add to 3. These numbers are 5 and -2 (since 5 × (-2) = -10 and 5 + (-2) = 3). So P(x) = (x + 5)(x - 2). From this factored form, we find zeros: (x + 5) = 0 → x = -5, and (x - 2) = 0 → x = 2. The zeros are x = -5, 2, giving x-intercepts (-5,0) and (2,0). Choice B correctly identifies x-intercepts as (-5,0), (2,0) by properly factoring the quadratic and applying the zero product property to each factor. Choice A incorrectly lists x-intercepts at (-10,0) and (1,0), possibly from misidentifying the factorization. When factoring x² + 3x - 10, we need factors of -10 that add to +3, which are 5 and -2, not -10 and 1. Always verify: (x + 5)(x - 2) = x² - 2x + 5x - 10 = x² + 3x - 10 ✓. 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!
For $P(x)=-x(x-3)(x+2)$ , identify the zeros and determine the correct end behavior to guide a rough sketch. Which option is correct?
Answer: Zeros: $x=0,3,-2$ ; end behavior: left up, right down 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 negative. Since 3 is odd and 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 high, right low. This end behavior plus the zeros gives us the skeleton of the graph! Choice B correctly identifies zeros as x=0, 3, -2 and end behavior left up, right down 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!
A polynomial is given by $P(x)=(x-3)(x+3)(x-1)^2.$ Use zeros, multiplicities, and degree to sketch a rough graph. Which statement is correct?
Answer: 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 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!
For $P(x)=-(x+1)(x-2)(x-4)$ , which description correctly matches the zeros and end behavior of the graph of $y=P(x)$ ?
Answer: Zeros at $x=-1,2,4$ ; as $x\to -\infty$ , $P(x)\to \infty$ and as $x\to \infty$ , $P(x)\to -\infty$ 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!
A polynomial is given by $P(x)=(x-3)(x+3)(x-1)^2.$ Use zeros, multiplicities, and degree to sketch a rough graph. Which statement is correct?
Answer: 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 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!
A quartic polynomial is given by $P(x)=(x-2)^2(x+1)(x+3).$ Use zeros and multiplicities to sketch a rough graph. Which description is correct?
Answer: Zeros at $x=2$ (mult. 2), $x=-1$ , $x=-3$ ; touches at $x=2$ , crosses at $x=-1$ and $x=-3$ ; both ends up 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!
Given the polynomial in factored form $P(x)=(x-3)(x+1)(x-2),$ identify the zeros of $P(x)$ and use them to sketch a rough graph. Your sketch should mark the x-intercepts and show the correct end behavior.
Answer: Zeros: $x=3,-1,2$ ; x-intercepts: $ (3,0), (-1,0), (2,0) $; end behavior: left down, right up; crosses at each zero. 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!
A polynomial profit model is $P(x)=(x-5)(x-1)(x+2).$ The zeros represent break-even points (where profit is $0$ ). Identify the break-even x-values and sketch a rough graph showing where $P(x)$ is positive or negative, including end behavior.
Answer: 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. 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!
A student factors the polynomial $p(x) = x^4 - 5x^2 + 4$ by first substituting $u = x^2$ to get $u^2 - 5u + 4 = (u - 1)(u - 4)$ . After substituting back, they obtain $p(x) = (x^2 - 1)(x^2 - 4)$ . How many x-intercepts does the graph of $y = p(x)$ have?
Answer: Four x-intercepts, because both $x^2 - 1 = (x-1)(x+1)$ and $x^2 - 4 = (x-2)(x+2)$ factor further 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.
Given the polynomial in factored form $P(x)=(x-1)(x+2)(x-3)$ , identify the zeros of $P(x)$ and use them to determine the x-intercepts of the graph.
Answer: Zeros: $x=1,-2,3$ ; x-intercepts: $(1,0),(-2,0),(3,0)$ 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+2)(x-3), we find zeros by setting each factor equal to zero: (x-1) = 0 → x = 1, (x+2) = 0 → x = -2, (x-3) = 0 → x = 3. The zeros are x = 1, -2, 3. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 3) gives zero at x = 3, and (x + 2) = (x - (-2)) gives zero at x = -2. Choice A correctly identifies zeros as x = 1, -2, 3 and shows x-intercepts at (1,0), (-2,0), (3,0) by properly applying the zero product property to each factor. Choice B has the signs wrong on all zeros: from the factor (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 + 2) 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!
Use the factored form $P(x)=x(x-2)^2(x+1)$ to answer: What are the x-intercepts, and how does the graph behave at each intercept (crosses vs touches)?
Answer: x-intercepts: $(-1,0)$ crosses, $(0,0)$ crosses, $(2,0)$ touches 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! Zeros at $x = 0$ and $x = -1$ have multiplicity 1 (odd), so the graph crosses at those points. Choice B correctly describes behavior as crosses at $x = -1$ and $x = 0$ , touches at $x = 2$ by recognizing multiplicity effects. Choice A shows the graph crossing at $x = 2$ , but this zero has even multiplicity 2, so the graph should touch and bounce at this point, not cross through. 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!
A company's profit is modeled by $P(x)=(x-1)(x-5)(x+2)$ , where zeros represent break-even points. Which set lists all break-even x-values (zeros) and the corresponding x-intercepts?
Answer: Break-even x-values: $-2,1,5$ ; x-intercepts: $(-2,0),(1,0),(5,0)$ 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!
For $P(x)=(x+3)^2(x-1),$ what are the x-intercepts and how does the graph behave at each intercept (crosses or touches)? Use this information to sketch a rough graph with correct end behavior.
Answer: x-intercepts: $(-3,0)$ and $(1,0)$ ; touches at $x=-3$ and crosses at $x=1$ ; end behavior: left down, right up. 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!
Given $P(x)=x(x-2)^2(x+1),$ identify the zeros (with multiplicities) and use them to sketch a rough graph. Indicate at which zeros the graph crosses the x-axis and at which it touches (bounces). Also state the end behavior.
Answer: 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. 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!
Use the factored form $P(x)=x(x-3)^2(x+2)$ to identify the zeros (with multiplicities) and state at which zeros the graph of $y=P(x)$ touches the x-axis rather than crossing it.
Answer: Touches at $x=3$ only 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! From the factored form P(x) = x(x-3)²(x+2), we find zeros by setting each factor equal to zero: x = 0 → x = 0 (multiplicity 1), (x-3)² = 0 → x = 3 (multiplicity 2), (x+2) = 0 → x = -2 (multiplicity 1). The zeros are x = 0, 3, -2. At x = 0 and x = -2, the multiplicity is 1 (odd), so the graph crosses. At x = 3, the multiplicity is 2 (even), so the graph touches and turns around without crossing. Choice C correctly states that the graph touches at x = 3 only by recognizing that this zero has even multiplicity 2 from the factor (x-3)². Choice D incorrectly claims the graph touches at x = 0 and x = -2, but these zeros have odd multiplicity 1, so the graph should cross at these points, not touch. 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!

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