Algebra Quiz: Factor Quadratics To Find Zeros

What this quiz covers

This quiz focuses on Factor Quadratics To Find Zeros, 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. $x=-4$ and $x=2$ (correct answer)
2. $x=4$ and $x=-2$
3. $x=-2$ and $x=4$
4. $x=-8$ and $x=1$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). The Zero Product Property is super useful: it tells us that if a product equals zero, then at least one of the factors must be zero. So if (x - 3)(x + 5) = 0, then either x - 3 = 0 (giving x = 3) or x + 5 = 0 (giving x = -5)—those are the zeros! Let's factor x² + 2x - 8: we need two numbers that multiply to -8 and add to 2. Those numbers are 4 and -2 because 4·-2 = -8 and 4 + -2 = 2. So the factored form is (x + 4)(x - 2). Now we use the Zero Product Property: set each factor to zero and solve. x + 4 = 0 gives x = -4, and x - 2 = 0 gives x = 2. Those are the zeros! Choice A is correct because it properly identifies the zeros as x = -4 and x = 2 using the Zero Product Property. Nicely done! Choice B gets the factors right but makes a mistake with the Zero Product Property: from (x + 4) = 0, we get x = -4, not x = 4. When you solve x + 4 = 0, you subtract 4 from both sides. The signs flip! Quick pattern to remember: in the factored form (x - p)(x - q), the zeros are x = p and x = q. Notice how the signs flip! If you have (x - 3), the zero is x = 3 (positive), and if you have (x + 5), the zero is x = -5 (negative). The sign in the factor is opposite the sign of the zero.

Question 2

1. $(3,0)$ and $(4,0)$
2. $(0,-3)$ and $(0,-4)$
3. $(-3,0)$ and $(-4,0)$ (correct answer)
4. $(-2,0)$ and $(-6,0)$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Zeros of a function are the x-values that make the function equal zero, and they're also where the graph crosses the x-axis (the x-intercepts). Finding zeros by factoring is often the fastest way, because once you have factors like (x - p)(x - q), the zeros are just x = p and x = q. Let's factor y = x² + 7x + 12: we need two numbers that multiply to 12 and add to 7. Those numbers are 3 and 4 because 3·4 = 12 and 3 + 4 = 7. So the factored form is (x + 3)(x + 4). Now we use the Zero Product Property: set each factor to zero and solve. x + 3 = 0 gives x = -3, and x + 4 = 0 gives x = -4. Those are the zeros! Choice A is correct because it properly shows the x-intercepts as points (-3, 0) and (-4, 0). Nicely done! Choice C gives the y-intercept or other value instead of the x-intercepts. Remember: zeros (x-intercepts) are where y = 0, which means where the graph crosses the x-axis, not the y-axis. To connect factoring to graphing: once you know the zeros, you know exactly where the parabola crosses the x-axis! If the zeros are x = -3 and x = -4, the graph goes through (-3, 0) and (-4, 0). The vertex is right in the middle at x = -3.5, and you can sketch the U-shape from there.

Question 3

1. $x=4$ and $x=4$
2. $x=1$ and $x=16$
3. $x=2$ and $x=8$ (correct answer)
4. $x=-2$ and $x=-8$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). To factor a quadratic like $x^2 + bx + c$, we need to find two numbers that multiply to give $c$ (the constant) and add to give $b$ (the coefficient of x), then write the factored form as $(x + \text{[first number]})(x + \text{[second number]})$. Let's factor $P(x) = -x^2 + 10x - 16$: First, factor out $-1$ to get $-(x^2 - 10x + 16)$. Now we need two numbers that multiply to $16$ and add to $-10$. Those numbers are $-2$ and $-8$ because $(-2) \cdot (-8) = 16$ and $(-2) + (-8) = -10$. So the factored form is $-(x - 2)(x - 8)$. Now we use the Zero Product Property: set each factor to zero and solve. $x - 2 = 0$ gives $x = 2$, and $x - 8 = 0$ gives $x = 8$. Those are the zeros! Choice A is correct because it properly identifies the zeros as $x = 2$ and $x = 8$ using the Zero Product Property. Nicely done! Choice B has a sign error in the factoring—a really common slip! When the constant is positive and the middle term is negative, we need two negative numbers that multiply to positive. It's easy to get signs mixed up, so always check by FOILing your factors back! Quick pattern to remember: in the factored form $(x - p)(x - q)$, the zeros are $x = p$ and $x = q$. Notice how the signs flip! If you have $(x - 3)$, the zero is $x = 3$ (positive), and if you have $(x + 5)$, the zero is $x = -5$ (negative). The sign in the factor is opposite the sign of the zero.

Question 4

1. $x = 0$ and $x = -5$
2. $x = -1$ and $x = 5$ (correct answer)
3. $x = 1$ and $x = -5$
4. $x = -5$ and $x = -1$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Zeros of a function are the x-values that make the function equal zero, and they're also where the graph crosses the x-axis (the x-intercepts). Finding zeros by factoring is often the fastest way, because once you have factors like (x - p)(x - q), the zeros are just x = p and x = q. Let's factor x² - 4x - 5: we need two numbers that multiply to -5 and add to -4. Those numbers are -5 and 1 because -5·1 = -5 and -5 + 1 = -4. So the factored form is (x - 5)(x + 1). Now we use the Zero Product Property: set each factor to zero and solve. x - 5 = 0 gives x = 5, and x + 1 = 0 gives x = -1. Those are the zeros! Choice A is correct because it properly identifies the zeros as x = -1 and x = 5 using the Zero Product Property. Nicely done! Choice B gets the factors right but makes a mistake with the Zero Product Property: from (x + 1) = 0, we get x = -1, not x = 1. When you solve x + 1 = 0, you subtract 1 from both sides. The signs flip! Quick pattern to remember: in the factored form (x - p)(x - q), the zeros are x = p and x = q. Notice how the signs flip! If you have (x - 3), the zero is x = 3 (positive), and if you have (x + 5), the zero is x = -5 (negative). The sign in the factor is opposite the sign of the zero. To connect factoring to graphing: once you know the zeros, you know exactly where the parabola crosses the x-axis! If the zeros are x = -1 and x = 5, the graph goes through (-1, 0) and (5, 0). The vertex is right in the middle at x = 2, and you can sketch the U-shape from there.

Question 5

1. $(x-3)(x-3)$; zeros: $x=3$
2. $(x+3)(x-3)$; zeros: $x=-3$ and $x=3$ (correct answer)
3. $(x+9)(x-1)$; zeros: $x=-9$ and $x=1$
4. $(x+3)(x+3)$; zeros: $x=-3$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). The Zero Product Property is super useful: it tells us that if a product equals zero, then at least one of the factors must be zero. So if (x - 3)(x + 5) = 0, then either x - 3 = 0 (giving x = 3) or x + 5 = 0 (giving x = -5)—those are the zeros! This is a difference of squares, which has a special pattern! For difference of squares: x² - 9 = x² - 3² = (x + 3)(x - 3), giving zeros at x = -3 and x = 3. Recognizing these patterns saves time! Choice B is correct because it properly factors the quadratic as (x + 3)(x - 3) and identifies the zeros as x = -3 and x = 3 using the Zero Product Property. Nicely done! Choice A only lists one zero when this quadratic actually has two! From (x + 3)(x - 3) = 0, we get both x = -3 AND x = 3. Don't forget to set each factor equal to zero! Watch out for perfect patterns: x² - 9 is a difference of squares = (x + 3)(x - 3) with zeros at ±3, and x² + 6x + 9 is a perfect square = (x + 3)² with one repeated zero at -3. Recognizing these patterns will speed you up!

Question 6

1. $x=3$ and $x=-4$ (correct answer)
2. $x=3$ and $x=4$
3. $x=-3$ and $x=-4$
4. $x=-3$ and $x=4$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Factoring reveals the zeros of a quadratic: the expression x² - 5x + 6 factors to (x - 2)(x - 3), which immediately tells us the zeros are x = 2 and x = 3 because those values make each factor (and therefore the whole expression) equal to zero. Let's factor x² + x - 12: we need two numbers that multiply to -12 and add to 1. Those numbers are 4 and -3 because 4·-3 = -12 and 4 + -3 = 1. So the factored form is (x + 4)(x - 3). Now we use the Zero Product Property: set each factor to zero and solve. x + 4 = 0 gives x = -4, and x - 3 = 0 gives x = 3. Those are the zeros! Choice B is correct because it properly identifies the zeros as x = 3 and x = -4 using the Zero Product Property. Nicely done! Choice A has a sign error in the factoring—a really common slip! When the constant is negative and the middle term is positive, we need one positive and one negative number, with the larger one positive. It's easy to get signs mixed up, so always check by FOILing your factors back! To check if x = 3 is really a zero, substitute it into the original: (3)² + (3) - 12 = 9 + 3 - 12 = 0. Yes! Since we get 0, this confirms x = 3 is indeed a zero. This is a great way to catch factoring mistakes!

Question 7

1. Zeros: $x=-3$ and $x=-\tfrac{1}{2}$ (correct answer)
2. Zeros: $x=3$ and $x=\tfrac{1}{2}$
3. Zeros: $x=-1$ and $x=-3$
4. Zeros: $x=-\tfrac{3}{2}$ and $x=-1$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). To factor a quadratic like x² + bx + c, we need to find two numbers that multiply to give c (the constant) and add to give b (the coefficient of x), then write the factored form as (x + [first number])(x + [second number]). For 2x² + 7x + 3, we can try grouping or trial: it factors to (2x + 1)(x + 3), since 2x·x = 2x², 2x·3 + 1·x = 7x, and 1·3 = 3. Now we use the Zero Product Property: set each factor to zero and solve. 2x + 1 = 0 gives x = -1/2, and x + 3 = 0 gives x = -3. Those are the zeros! Choice A is correct because it properly factors the quadratic as (2x + 1)(x + 3) and identifies the zeros as x = -3 and x = -1/2 using the Zero Product Property. Nicely done! Choice B gets the factors right but makes a mistake with the Zero Product Property: from (x + 3) = 0, we get x = -3, not x = 3. When you solve x + 3 = 0, you subtract 3 from both sides. The signs flip! Quick pattern to remember: in the factored form (x - p)(x - q), the zeros are x = p and x = q. Notice how the signs flip! If you have (x - 3), the zero is x = 3 (positive), and if you have (x + 5), the zero is x = -5 (negative). The sign in the factor is opposite the sign of the zero.

Question 8

1. $x=1$ and $x=5$ (correct answer)
2. $x=-1$ and $x=-5$
3. $x=2$ and $x=3$
4. $x=0$ and $x=5$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Factoring reveals the zeros of a quadratic: the expression x² - 5x + 6 factors to (x - 2)(x - 3), which immediately tells us the zeros are x = 2 and x = 3 because those values make each factor (and therefore the whole expression) equal to zero. Let's factor x² - 6x + 5: we need two numbers that multiply to 5 and add to -6. Those numbers are -1 and -5 because -1·-5 = 5 and -1 + -5 = -6. So the factored form is (x - 1)(x - 5). Now we use the Zero Product Property: set each factor to zero and solve. x - 1 = 0 gives x = 1, and x - 5 = 0 gives x = 5. Those are the zeros! Choice A is correct because it properly identifies the zeros as x = 1 and x = 5 using the Zero Product Property. Nicely done! Choice B has a sign error in the factoring—a really common slip! When the constant is positive and the middle term is negative, we need both numbers negative. It's easy to get signs mixed up, so always check by FOILing your factors back! The Zero Product Property is your friend: whenever you have something like (x - 3)(x + 5) = 0, you know that one of those factors must be zero, so x - 3 = 0 OR x + 5 = 0. Solve each little equation separately, and boom—you've got your zeros! They're x = 3 and x = -5.

Question 9

1. Zeros: $x = 0$ and $x = -9$
2. Zeros: $x = -9$ and $x = 1$
3. Zero: $x = -3$ (correct answer)
4. Zeros: $x = -3$ and $x = 3$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Factoring reveals the zeros of a quadratic: the expression x² + 6x + 9 factors to (x + 3)², which immediately tells us the zero is x = -3 (repeated) because that value makes the factor (and therefore the whole expression) equal to zero. This is a perfect square trinomial, which has a special pattern! For perfect square: x² + 6x + 9 = (x + 3)², giving one zero (repeated) at x = -3. Recognizing these patterns saves time! To check if x = -3 is really a zero, substitute it into the original: (-3)² + 6(-3) + 9 = 9 - 18 + 9 = 0. Yes! Since we get 0, this confirms x = -3 is indeed a zero. This is a great way to catch factoring mistakes! Choice A is correct because it properly identifies the zero as x = -3. Nicely done! Choice C only lists one zero when this quadratic actually has one (repeated), but it incorrectly suggests two distinct zeros! From (x + 3)² = 0, we get x = -3 (twice). Don't forget it's repeated! Watch out for perfect patterns: x² - 9 is a difference of squares = (x + 3)(x - 3) with zeros at ±3, and x² + 6x + 9 is a perfect square = (x + 3)² with one repeated zero at -3. Recognizing these patterns will speed you up! To connect factoring to graphing: once you know the zeros, you know exactly where the parabola crosses the x-axis! If the zero is x = -3 (repeated), the graph touches the x-axis at (-3, 0) and bounces back, like a U-shape just grazing the axis.

Question 10

1. -17
2. 17
3. 13 (correct answer)
4. -13

Explanation: Expand $(x + 3)(x - 5)$: $(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$. Comparing with$f(x) = x^2 + bx + c$, we have$b = -2$and$c = -15$. Therefore$b - c = (-2) - (-15) = -2 + 15 = 13$. Choice A results from calculating$c - b$instead of$b - c$. Choice B comes from sign errors in the expansion. Choice D results from incorrectly computing$b - c$as$-2 - 15 = -17$$ and then making a sign error.

Question 11

1. 0 (correct answer)
2. 1
3. 2
4. 4

Explanation: Factor $f(x) = x^2 - 8x + 15$. We need two numbers that multiply to 15 and add to 8. These are 3 and 5. So $f(x) = (x - 3)(x - 5)$, giving zeros $x = 3$ and $x = 5$. Factor $g(x) = x^2 - 8x + 16$. We need two numbers that multiply to 16 and add to 8. These are 4 and 4. So $g(x) = (x - 4)(x - 4) = (x - 4)^2$, giving a double zero at $x = 4$. The zeros of $f(x)$ are 3 and 5. The zero of $g(x)$ is 4. These functions share no common zeros. Choice B suggests they share one zero, which would occur if one of 3, 5 equaled 4. Choice C suggests they share two zeros, which is impossible since $g(x)$ has only one distinct zero. Choice D is impossible since there are only 3 distinct zeros total between both functions.

Question 12

1. $x = -6$ and $x = 4$
2. $x = -3$ and $x = 2$ (correct answer)
3. $x = -1.5$ and $x = 1$
4. $x = 0$ and $x = -1$

Explanation: When a quadratic expression is multiplied by a constant, the zeros remain the same. This is because if $f(x) = 0$ when $x = a$, then $2f(x) = 2 \cdot 0 = 0$ when $x = a$. The original expression $x^2 + px + q$ has zeros at $x = -3$ and $x = 2$, so the new expression $2x^2 + 2px + 2q = 2(x^2 + px + q)$ also has zeros at $x = -3$ and $x = 2$. Choice A results from incorrectly doubling the zeros. Choice C comes from incorrectly halving the zeros. Choice D results from completely misunderstanding the relationship between scaling and zeros.

Question 13

1. 12 (correct answer)
2. 16
3. 20
4. 24

Explanation: First factor out the GCF: $3x^2 - 21x + 36 = 3(x^2 - 7x + 12)$. Now factor $x^2 - 7x + 12$. We need two numbers that multiply to 12 and add to 7. These are 3 and 4. So $x^2 - 7x + 12 = (x - 3)(x - 4)$. Therefore $3x^2 - 21x + 36 = 3(x - 3)(x - 4)$. We can write this as $(x - 3) \cdot 3(x - 4)$, so $a = 3$ and $b = 4$. Therefore $a \cdot b = 3 \cdot 4 = 12$. Choice B results from using the wrong factorization. Choice C comes from adding $a$ and $b$ instead of multiplying. Choice D results from calculation errors in the factoring process.

Question 14

1. 1
2. 9
3. 3
4. 5 (correct answer)

Explanation: When you encounter a quadratic expression and need to find information about its zeros, your first step should be to factor the expression completely. This reveals the zeros directly and makes further calculations much easier. Let's factor $4x^2 - 4x - 8$. First, notice that all terms share a common factor of 4, so we can factor that out: $4(x^2 - x - 2)$. Now we need to factor the quadratic $x^2 - x - 2$. We're looking for two numbers that multiply to -2 and add to -1. Those numbers are -2 and 1, giving us $x^2 - x - 2 = (x - 2)(x + 1)$. So the complete factorization is $4(x - 2)(x + 1)$. The zeros occur when the expression equals zero, which happens when $x - 2 = 0$ or $x + 1 = 0$. Therefore, the zeros are $x = 2$ and $x = -1$. The sum of the squares of the zeros is $2^2 + (-1)^2 = 4 + 1 = 5$. Looking at the wrong answers: A) 1 would be just the square of one zero ($(-1)^2$), missing the other. B) 9 might come from incorrectly calculating $(2 + (-1))^2 = 1^2 = 1$ then making an arithmetic error, or from squaring 3 if you mistakenly thought the sum of zeros was 3. C) 3 could result from adding the zeros first ($2 + (-1) = 1$) then adding 2, confusing the sum of zeros with sum of squares. Always factor quadratics completely to find zeros clearly, then carefully apply the operation requested—whether it's sum, product, or sum of squares.

Question 15

1. Returns at $t = 16$ seconds, total air time is 16 seconds
2. Returns at $t = 2$ seconds, total air time is 2 seconds
3. Returns at $t = 8$ seconds, total air time is 8 seconds
4. Returns at $t = 4$ seconds, total air time is 4 seconds (correct answer)

Explanation: When you encounter a quadratic function representing projectile motion, you're looking for when the object hits the ground—which occurs when the height equals zero. Finding the zeros through factoring will give you the complete picture of the ball's flight. To find when $h(t) = 0$, set the equation equal to zero: $-16t^2 + 64t = 0$. Factor out the common term $-16t$: $-16t(t - 4) = 0$. Using the zero product property, either $-16t = 0$ or $t - 4 = 0$, giving you $t = 0$ and $t = 4$. These zeros tell the complete story: $t = 0$ is when the ball is thrown (starts at ground level), and $t = 4$ is when it returns to the ground. The total air time is $4 - 0 = 4$ seconds. Choice A incorrectly uses the coefficient 16 from the equation as the time value, confusing the mathematical constants with the physical meaning. Choice B likely comes from incorrectly factoring or misunderstanding that you need both zeros to determine the flight duration. Choice C might result from calculation errors when factoring, perhaps confusing the vertex time (when the ball reaches maximum height) with the landing time. The correct answer is D: the ball returns to the ground at $t = 4$ seconds, with a total air time of 4 seconds. Strategy tip: For projectile motion problems, always factor completely to find both zeros—they represent the start and end of the flight, giving you the total time in the air.

Question 16

1. $x = -3$ and $x = 7$
2. $x = -6$ and $x = 7$
3. $x = -7$ and $x = 3$ (correct answer)
4. $x = -14$ and $x = 6$

Explanation: When you encounter a quadratic expression and need to find its zeros, you're looking for the x-values where the function equals zero. The most efficient approach is to factor the expression first, then use the zero product property. To factor $2x^2 + 8x - 42$, start by factoring out the greatest common factor of 2: $2(x^2 + 4x - 21)$. Now you need to factor the trinomial $x^2 + 4x - 21$. Look for two numbers that multiply to -21 and add to 4. Those numbers are 7 and -3, since $7 \times (-3) = -21$ and $7 + (-3) = 4$. So the complete factorization is $2(x + 7)(x - 3)$. To find the zeros, set this equal to zero: $2(x + 7)(x - 3) = 0$. Since 2 ≠ 0, you need either $(x + 7) = 0$ or $(x - 3) = 0$. This gives you $x = -7$ and $x = 3$. Choice A gives $x = -3$ and $x = 7$, which reverses the signs from the correct zeros. Choice B includes $x = -6$, which doesn't satisfy the original equation. Choice D provides $x = -14$ and $x = 6$, which are completely incorrect values that don't work when substituted back into the original expression. Remember this key pattern: when factoring reveals $(x + a)(x - b) = 0$, the zeros are $x = -a$ and $x = b$. Always double-check by substituting your zeros back into the original expression to verify they make it equal zero.

Question 17

1. $x = 5$ and $x = 7$ and all values between them
2. $x = 5$ and $x = 7$ only (correct answer)
3. $x = -5$ and $x = -7$ only
4. $x = 35$ and $x = 1$ only

Explanation: When you encounter a profit function and need to find break-even points, you're looking for where the profit equals zero. Break-even means the business makes neither profit nor loss, so you need to solve $P(x) = 0$. To find where $P(x) = -x^2 + 12x - 35 = 0$, you can factor this quadratic equation. Look for two numbers that multiply to $-35$ and add to $12$. Since $-35 = -5 \times 7$ and $-5 + 7 = 2$, this doesn't work directly. Instead, notice that $-35 = 5 \times (-7)$ and we need the middle term to be $+12x$. Rearranging: $-x^2 + 12x - 35 = -(x^2 - 12x + 35) = -(x - 5)(x - 7) = 0$ This gives us $x = 5$ and $x = 7$ as the solutions. Choice B correctly identifies these two specific values where profit equals zero. Choice A incorrectly suggests the business breaks even for all values between 5 and 7, but break-even points are specific values, not ranges. Since this is a downward-opening parabola, the business actually makes a profit between $x = 5$ and $x = 7$. Choice C gives negative values, which don't make sense in this context since you can't sell a negative number of items. Choice D provides completely incorrect values that don't satisfy the equation. Remember: break-even points are the exact $x$-intercepts of a profit function, not intervals. Always check that your solutions make sense in the real-world context of the problem.

Question 18

1. $(2, 0)$ and $(3, 0)$ (correct answer)
2. $(-2, 0)$ and $(-3, 0)$
3. $(0, 2)$ and $(0, 3)$
4. $(1, 0)$ and $(6, 0)$

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Zeros of a function are the x-values that make the function equal zero, and they're also where the graph crosses the x-axis (the x-intercepts). Finding zeros by factoring is often the fastest way, because once you have factors like (x - p)(x - q), the zeros are just x = p and x = q. Let's factor x² - 5x + 6: we need two numbers that multiply to 6 and add to -5. Those numbers are -2 and -3 because -2·-3 = 6 and -2 + -3 = -5. So the factored form is (x - 2)(x - 3). Now we use the Zero Product Property: set each factor to zero and solve. x - 2 = 0 gives x = 2, and x - 3 = 0 gives x = 3, so the x-intercepts are (2, 0) and (3, 0). Choice A is correct because it properly identifies the zeros as x = 2 and x = 3 using the Zero Product Property and shows the x-intercepts as points (2, 0) and (3, 0). Nicely done! Choice C has the x-intercepts as (0, y) instead of (x, 0)—the coordinates are backwards! X-intercepts are always in the form (x-value, 0), not (0, x-value). To connect factoring to graphing: once you know the zeros, you know exactly where the parabola crosses the x-axis! If the zeros are x = 2 and x = 3, the graph goes through (2, 0) and (3, 0). The vertex is right in the middle at x = 2.5, and you can sketch the U-shape from there.

Question 19

1. Zeros: $x=9$ and $x=-9$
2. Zeros: $x=3$ and $x=-3$ (correct answer)
3. Zeros: $x=0$ and $x=9$
4. Zeros: $x=3$ only

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). Factoring reveals the zeros of a quadratic: the expression x² - 5x + 6 factors to (x - 2)(x - 3), which immediately tells us the zeros are x = 2 and x = 3 because those values make each factor (and therefore the whole expression) equal to zero. This is a difference of squares, which has a special pattern! For difference of squares: x² - a² = (x + a)(x - a), giving zeros at x = -a and x = a. Here, x² - 9 = x² - 3² = (x + 3)(x - 3), giving zeros at x = -3 and x = 3. Recognizing these patterns saves time! Choice B is correct because it properly factors the quadratic as (x + 3)(x - 3) and identifies the zeros as x = 3 and x = -3 using the Zero Product Property. Nicely done! Choice A found two numbers that add to 0 but don't multiply to -9—close! Remember, for x² + bx + c, we need numbers that do BOTH: multiply to c AND add to b. In this case, 9 and -9 multiply to -81, not -9. Watch out for perfect patterns: x² - 9 is a difference of squares = (x + 3)(x - 3) with zeros at ±3, and x² + 6x + 9 is a perfect square = (x + 3)² with one repeated zero at -3. Recognizing these patterns will speed you up!

Question 20

1. $t=-1$ and $t=-3$
2. $t=0$ and $t=3$
3. $t=1$ only
4. $t=1$ and $t=3$ (correct answer)

Explanation: This question tests your understanding of factoring quadratic expressions and using the Zero Product Property to find zeros (the x-intercepts where the graph crosses the x-axis). The Zero Product Property is super useful: it tells us that if a product equals zero, then at least one of the factors must be zero. So if (x - 3)(x + 5) = 0, then either x - 3 = 0 (giving x = 3) or x + 5 = 0 (giving x = -5)—those are the zeros! To make factoring easier, first multiply h(t) by -1 to get t² - 4t + 3 (the zeros will be the same). We need two numbers that multiply to 3 and add to -4. Those numbers are -1 and -3 because -1·-3 = 3 and -1 + -3 = -4. So the factored form is (t - 1)(t - 3). Now we use the Zero Product Property: set each factor to zero and solve. t - 1 = 0 gives t = 1, and t - 3 = 0 gives t = 3. Those are the zeros! Choice A is correct because it properly identifies the zeros as t = 1 and t = 3 using the Zero Product Property. Nicely done! Choice B has a sign error in the factoring—a really common slip! When the constant is positive and the middle term is negative, we need both numbers negative. It's easy to get signs mixed up, so always check by FOILing your factors back! The Zero Product Property is your friend: whenever you have something like (x - 3)(x + 5) = 0, you know that one of those factors must be zero, so x - 3 = 0 OR x + 5 = 0. Solve each little equation separately, and boom—you've got your zeros! They're x = 3 and x = -5.