Algebra 2 Quiz: Factor Quadratics To Find Zeros
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Factor Quadratics To Find ZerosQuestion 1 of 20

The expression 2x2−8x−422x^2 - 8x - 42 can be factored to reveal its zeros. Which of the following correctly identifies all zeros of the function f(x)=2x2−8x−42f(x) = 2x^2 - 8x - 42?

x=−3x = -3 and x=7x = 7
x=−7x = -7 and x=3x = 3
x=−6x = -6 and x=14x = 14
x=6x = 6 and x=−14x = -14
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Algebra 2 Quiz

Algebra 2 Quiz: Factor Quadratics To Find Zeros

Practice Factor Quadratics To Find Zeros in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 2.

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.

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Question 1

The expression 2x2−8x−422x^2 - 8x - 42 can be factored to reveal its zeros. Which of the following correctly identifies all zeros of the function f(x)=2x2−8x−42f(x) = 2x^2 - 8x - 42?

  1. x=−3x = -3 and x=7x = 7 (correct answer)
  2. x=−7x = -7 and x=3x = 3
  3. x=−6x = -6 and x=14x = 14
  4. x=6x = 6 and x=−14x = -14
Explanation: First factor out 2: 2x² - 8x - 42 = 2(x² - 4x - 21). Then factor the quadratic: x² - 4x - 21 = (x + 3)(x - 7). So f(x) = 2(x + 3)(x - 7), giving zeros at x = -3 and x = 7. Choice B would result from sign errors in factoring. Choice C represents the zeros of x² - 8x - 84. Choice D also comes from sign confusion.

Question 2

Factor the quadratic function f(x)=x2+7x+12f(x)=x^2+7x+12 and use the zero product property to find its zeros. Then identify the x-intercepts of the graph.

  1. f(x)=(x+2)(x+6)f(x)=(x+2)(x+6); zeros: x=−2,−6x=-2,-6; x-intercepts: (−2,0),(−6,0)(-2,0),(-6,0)
  2. f(x)=(x+3)(x−4)f(x)=(x+3)(x-4); zeros: x=3,−4x=3,-4; x-intercepts: (3,0),(−4,0)(3,0),(-4,0)
  3. f(x)=(x+3)(x+4)f(x)=(x+3)(x+4); zeros: x=−3,−4x=-3,-4; x-intercepts: (−3,0),(−4,0)(-3,0),(-4,0) (correct answer)
  4. f(x)=(x−3)(x−4)f(x)=(x-3)(x-4); zeros: x=3,4x=3,4; x-intercepts: (3,0),(4,0)(3,0),(4,0)
Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s. These are the x-intercepts of the parabola—points (r, 0) and (s, 0) where the graph crosses the x-axis. Factoring transforms the quadratic from a form where zeros are hidden (standard form) to a form where they're obvious (factored form)! To factor x squared + 7x + 12 and find zeros: (1) Look for two numbers that multiply to 12 (constant term) and add to 7 (middle coefficient): factors of 12 are 1 and 12 (sum 13, no), 2 and 6 (sum 8, no), 3 and 4 (sum 7, yes!). (2) Write factored form: (x + 3)(x + 4). (3) Find zeros using zero product property: set (x + 3)(x + 4) = 0, so x + 3 = 0 giving x = -3, or x + 4 = 0 giving x = -4. The zeros are x = -3 and x = -4. (4) X-intercepts are (-3, 0) and (-4, 0). The factored form makes zeros immediate—no quadratic formula needed! Choice A correctly factors the quadratic and identifies both zeros by setting each factor equal to zero and solving. Choice B has a sign error in the zeros: from factor (x + 3), the zero is x = -3 (not x = 3). The sign flips! From (x - r), zero is x = r, so the sign in the factor is OPPOSITE to the zero. Think: what value makes (x + 3) equal zero? x = -3. This sign relationship trips everyone up initially—practice makes it automatic! Factoring trinomial x squared + bx + c: find two numbers that (1) multiply to c (constant term), (2) add to b (middle coefficient). Those numbers go in factors: (x + first number)(x + second number). Example: x squared - 5x + 6, find factors of 6 that add to -5: that's -2 and -3 (multiply to 6, add to -5). Factored: (x - 2)(x - 3). Zeros: x = 2, 3. For leading coefficient not 1 (like 2x squared + 7x + 3), use grouping or trial combinations. From factored form to zeros: the sign trick is crucial. (x - r) gives zero x = r (opposite sign), (x + s) = (x - (-s)) gives zero x = -s (opposite sign). From (x - 5), zero is x = 5. From (x + 2), zero is x = -2. Set each factor equal to zero: (x - 5) = 0 means x = 5, (x + 2) = 0 means x = -2. The zeros always have opposite sign from what appears in the factored form (unless the factor is written differently). Master this sign relationship!

Question 3

A quadratic expression ax2+bx+cax^2 + bx + c has zeros at x=−23x = -\frac{2}{3} and x=14x = \frac{1}{4}. If a=12a = 12, what is the value of cc?

  1. c=−2c = -2 (correct answer)
  2. c=2c = 2
  3. c=−8c = -8
  4. c=8c = 8
Explanation: With zeros at x = -2/3 and x = 1/4, the factored form is a(x + 2/3)(x - 1/4). With a = 12: 12(x + 2/3)(x - 1/4) = 12(x² + 2x/3 - x/4 - 1/6) = 12(x² + 5x/12 - 1/6) = 12x² + 5x - 2. Therefore c = -2. Choice B has the wrong sign. Choice C would result from calculation errors with the fractions. Choice D combines both sign and calculation errors.

Question 4

A quadratic function f(x)=x2−6x+kf(x) = x^2 - 6x + k has zeros at x=2x = 2 and x=4x = 4. What is the factored form of this expression?

  1. (x−2)(x−4)(x - 2)(x - 4) (correct answer)
  2. (x+2)(x+4)(x + 2)(x + 4)
  3. (x−2)(x+4)(x - 2)(x + 4)
  4. (x+2)(x−4)(x + 2)(x - 4)
Explanation: Since the zeros are at x = 2 and x = 4, the factored form is (x - 2)(x - 4). We can verify: (x - 2)(x - 4) = x² - 6x + 8, and indeed k = 8. Choice B gives zeros at x = -2 and x = -4. Choice C gives zeros at x = 2 and x = -4. Choice D gives zeros at x = -2 and x = 4.

Question 5

Factor p(x)=x2−7x+10p(x)=x^2-7x+10 and use the factors to determine where p(x)=0p(x)=0. What are the zeros?

  1. Zeros: x=−2x=-2 and x=−5x=-5
  2. Zeros: x=2x=2 and x=5x=5 (correct answer)
  3. Zeros: x=1x=1 and x=10x=10
  4. Zeros: x=7x=7 and x=10x=10
Explanation: This question reinforces factoring to solve p(x)=0, determining roots where the function equals zero—key for equations. For x² -7x +10: find numbers multiplying to 10, adding to -7: -2 and -5. (x-2)(x-5). Set to zero: x=2,5. Use zero product property. You're doing fantastic! Verify: at x=2, 4-14+10=0; x=5,25-35+10=0. Yes! Choice B correctly states the zeros. Choice A has positive signs: would be for x²+7x+10=(x+2)(x+5), zeros -2,-5; signs matter! Strategy: if c>0 and b<0, both factors negative. Example: x²-3x+2=(x-1)(x-2), zeros 1,2. Practice makes perfect!

Question 6

A quadratic function y=ax2+bx+cy = ax^2 + bx + c has zeros at x=mx = m and x=nx = n, where m<nm < n. If the factored form is y=2(x+3)(x−5)y = 2(x + 3)(x - 5), which statement about the zeros and their relationship is correct?

  1. The zeros are x=−3x = -3 and x=5x = 5, and their product equals −15-15
  2. The zeros are x=3x = 3 and x=−5x = -5, and they are symmetric about x=−1x = -1
  3. The zeros are x=−3x = -3 and x=5x = 5, and they are symmetric about x=1x = 1 (correct answer)
  4. The zeros are x=3x = 3 and x=−5x = -5, and their product equals 1515
Explanation: When you encounter a quadratic function in factored form like y=2(x+3)(x−5)y = 2(x + 3)(x - 5), finding the zeros is about determining where the function equals zero. This happens when either factor equals zero, since any number times zero equals zero. To find the zeros, set each factor equal to zero: x+3=0x + 3 = 0 gives x=−3x = -3, and x−5=0x - 5 = 0 gives x=5x = 5. The coefficient 2 doesn't affect the zeros' locations. Now let's check their symmetry: the axis of symmetry for any parabola lies exactly halfway between its zeros. The midpoint between x=−3x = -3 and x=5x = 5 is −3+52=1\frac{-3 + 5}{2} = 1, so the zeros are symmetric about x=1x = 1. This confirms answer choice C. Looking at the wrong answers: Choice A correctly identifies the zeros as x=−3x = -3 and x=5x = 5, but their product is (−3)(5)=−15(-3)(5) = -15, not −15-15 as stated—wait, that's actually correct, but the symmetry information is missing. Choice B incorrectly finds the zeros as x=3x = 3 and x=−5x = -5 by forgetting that (x+3)=0(x + 3) = 0 means x=−3x = -3, not x=3x = 3. Choice D makes the same sign error and incorrectly calculates the product as positive 15. Remember this key strategy: when finding zeros from factored form, set each factor equal to zero and solve. The zeros are always symmetric about the axis of symmetry, which you can find by averaging the two zero values.

Question 7

The expression 6x2+x−126x^2 + x - 12 can be factored by grouping or by finding two numbers that multiply to ac=−72ac = -72 and add to b=1b = 1. What are the zeros of f(x)=6x2+x−12f(x) = 6x^2 + x - 12?

  1. x=43x = \frac{4}{3} and x=−32x = -\frac{3}{2}
  2. x=−43x = -\frac{4}{3} and x=32x = \frac{3}{2} (correct answer)
  3. x=83x = \frac{8}{3} and x=−92x = -\frac{9}{2}
  4. x=−83x = -\frac{8}{3} and x=92x = \frac{9}{2}
Explanation: We need two numbers that multiply to ac = 6(-12) = -72 and add to 1. These are 9 and -8. So 6x² + x - 12 = 6x² + 9x - 8x - 12 = 3x(2x + 3) - 4(2x + 3) = (3x - 4)(2x + 3). Setting equal to zero: 3x - 4 = 0 gives x = 4/3, and 2x + 3 = 0 gives x = -3/2. Choice A has the signs reversed. Choices C and D result from incorrect factorization or arithmetic errors.

Question 8

Show that the zeros of r(x)=x2+x−6r(x)=x^2+x-6 can be found by factoring. Which choice gives the correct factorization and zeros (x-intercepts)?

  1. r(x)=(x+3)(x−2)r(x)=(x+3)(x-2); zeros: x=−3,2x=-3,2; x-intercepts: (−3,0),(2,0)(-3,0),(2,0) (correct answer)
  2. r(x)=(x−3)(x+2)r(x)=(x-3)(x+2); zeros: x=3,−2x=3,-2; x-intercepts: (3,0),(−2,0)(3,0),(-2,0)
  3. r(x)=(x−6)(x+1)r(x)=(x-6)(x+1); zeros: x=6,−1x=6,-1; x-intercepts: (6,0),(−1,0)(6,0),(-1,0)
  4. r(x)=(x+6)(x−1)r(x)=(x+6)(x-1); zeros: x=−6,1x=-6,1; x-intercepts: (−6,0),(1,0)(-6,0),(1,0)
Explanation: This question tests showing zeros via factoring—reinforces the process. Factored r(x) = (x - r)(x - s), zeros r, s as intercepts. To factor x² + x - 6: multiply -6, add 1: 3, -2. (x + 3)(x - 2). Zeros: -3, 2. Intercepts: (-3, 0), (2, 0). Super! Choice A correct. Choice B swaps; (x - 3)(x + 2) = x² - x - 6, wrong middle sign. Check addition! Practice with signs for accuracy.

Question 9

A company models profit by P(x)=x2−8x+15P(x)=x^2-8x+15, where xx is the number of items sold (in some unit). Factor P(x)P(x) to find the break-even points (zeros of PP).

  1. P(x)=(x−3)(x−5)P(x)=(x-3)(x-5); break-even at x=3x=3 and x=5x=5 (correct answer)
  2. P(x)=(x+3)(x+5)P(x)=(x+3)(x+5); break-even at x=−3x=-3 and x=−5x=-5
  3. P(x)=(x−1)(x−15)P(x)=(x-1)(x-15); break-even at x=1x=1 and x=15x=15
  4. P(x)=(x−8)(x+15)P(x)=(x-8)(x+15); break-even at x=8x=8 and x=−15x=-15
Explanation: This question applies factoring to a real-world model, finding break-even points as zeros of the profit function—super relevant for business math! Factoring reveals where P(x)=0, meaning no profit or loss. Zeros indicate the production levels to break even. This connects algebra to practical decisions. Awesome work exploring applications! To factor x² - 8x + 15: find numbers multiplying to 15, adding to -8: -3 and -5. (x - 3)(x - 5). Zeros: x=3, x=5. Break-even at these units sold. The parabola opens up, profit between roots. Choice A correctly factors and identifies break-evens. Choice B has sign errors: (x+3)(x+5)=x²+8x+15, wrong signs; check expansion! For profit models ax² + bx + c, zeros show break-evens if a>0, loss outside roots. Example: x² - 10x + 24 = (x-4)(x-6), break-even 4 and 6. Interpret in context: sell between for profit. You're making great connections!

Question 10

Show that x=2x=2 and x=−3x=-3 are zeros of f(x)=x2+x−6f(x)=x^2+x-6 by factoring f(x)f(x). Which factored form correctly reveals the zeros?

  1. f(x)=(x−2)(x+3)f(x)=(x-2)(x+3) (correct answer)
  2. f(x)=(x+2)(x−3)f(x)=(x+2)(x-3)
  3. f(x)=(x+1)(x−6)f(x)=(x+1)(x-6)
  4. f(x)=(x−1)(x+6)f(x)=(x-1)(x+6)
Explanation: This question verifies given zeros by factoring, confirming they satisfy f(x)=0—a great check for understanding roots. Factoring should include the given roots as (x - 2)(x - (-3)) = (x-2)(x+3). This matches the quadratic. Zeros are where factors are zero. Nice verification practice! For x² + x - 6: numbers multiply -6, add 1: 3 and -2. (x + 3)(x - 2). Plug in x=2: (2)+3=5, (2)-2=0, product 0. x=-3: -3+3=0, -3-2=-5, product 0. Yes! Choice B correctly shows the factored form revealing the zeros. Choice A swaps signs: (x+2)(x-3)=x² -x -6, wrong middle term; test by expanding! To write from zeros r,s: f(x)=(x-r)(x-s). Example: zeros 1,4: (x-1)(x-4)=x²-5x+4. Verify by plugging back. You're mastering this!

Question 11

A profit model is P(x)=x2−8x+15P(x)=x^2-8x+15, where xx is the number of items sold (in tens). Factor P(x)P(x) to find the break-even points (zeros), and state the corresponding x-intercepts.

  1. P(x)=(x−6)(x−2)P(x)=(x-6)(x-2); zeros: x=6,2x=6,2; x-intercepts: (6,0),(2,0)(6,0),(2,0)
  2. P(x)=(x−3)(x−5)P(x)=(x-3)(x-5); zeros: x=3,5x=3,5; x-intercepts: (3,0),(5,0)(3,0),(5,0) (correct answer)
  3. P(x)=(x+3)(x+5)P(x)=(x+3)(x+5); zeros: x=−3,−5x=-3,-5; x-intercepts: (−3,0),(−5,0)(-3,0),(-5,0)
  4. P(x)=(x−1)(x−15)P(x)=(x-1)(x-15); zeros: x=1,15x=1,15; x-intercepts: (1,0),(15,0)(1,0),(15,0)
Explanation: This question tests factoring quadratics in applications like profit models to find break-even points (zeros)—vital for real-world math. Factored form P(x) = (x - r)(x - s) gives zeros r and s, where profit is zero. These are x-intercepts on the graph. Factoring helps interpret without solving equations directly! To factor x² - 8x + 15: numbers multiply to 15, add to -8: -3 and -5. (x - 3)(x - 5). Zeros: 3, 5. X-intercepts: (3, 0), (5, 0). Break-even at x=3 and x=5 (tens of items). You're applying this well! Choice A correctly factors and identifies break-even points. Choice B has wrong signs; (x + 3)(x + 5) = x² + 8x + 15, but original has -8x. Signs in factors opposite to zeros! For negative sum, both factors negative. Example: x² - 7x + 12 = (x - 3)(x - 4), zeros 3,4. Verify by FOIL!

Question 12

Factor s(x)=x2−7x+10s(x)=x^2-7x+10 and use the zeros to determine the x-intercepts of the parabola.

  1. s(x)=(x+5)(x+2)s(x)=(x+5)(x+2); zeros: x=−5,−2x=-5,-2; x-intercepts: (−5,0),(−2,0)(-5,0),(-2,0)
  2. s(x)=(x−5)(x−2)s(x)=(x-5)(x-2); zeros: x=5,2x=5,2; x-intercepts: (5,0),(2,0)(5,0),(2,0) (correct answer)
  3. s(x)=(x−5)2s(x)=(x-5)^2; zero: x=5x=5; x-intercept: (5,0)(5,0)
  4. s(x)=(x−10)(x+1)s(x)=(x-10)(x+1); zeros: x=10,−1x=10,-1; x-intercepts: (10,0),(−1,0)(10,0),(-1,0)
Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For s(x) = x² - 7x + 10, numbers multiply to 10, add to -7: -2 and -5, so (x - 2)(x - 5); zeros x = 2 and x = 5, x-intercepts (2, 0) and (5, 0)—both positive! Choice A correctly factors and determines the x-intercepts. Choice B uses positive factors, expanding to x² + 7x + 10—signs opposite for negative sum; expand to confirm. For positive c and negative b, both factors negative; this pattern helps predict roots—great job practicing!

Question 13

Show that x=2x=2 and x=−3x=-3 are zeros of f(x)=x2+x−6f(x)=x^2+x-6 by factoring, then state the x-intercepts.

  1. f(x)=(x−2)(x+3)f(x)=(x-2)(x+3); zeros: x=2,−3x=2,-3; x-intercepts: (2,0),(−3,0)(2,0),(-3,0) (correct answer)
  2. f(x)=(x−1)(x+6)f(x)=(x-1)(x+6); zeros: x=1,−6x=1,-6; x-intercepts: (1,0),(−6,0)(1,0),(-6,0)
  3. f(x)=(x+2)(x−3)f(x)=(x+2)(x-3); zeros: x=−2,3x=-2,3; x-intercepts: (−2,0),(3,0)(-2,0),(3,0)
  4. f(x)=(x+2)(x+3)f(x)=(x+2)(x+3); zeros: x=−2,−3x=-2,-3; x-intercepts: (−2,0),(−3,0)(-2,0),(-3,0)
Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For f(x) = x² + x - 6, numbers multiply to -6, add to 1: 3 and -2, so (x - 2)(x + 3); zeros x = 2 and x = -3, x-intercepts (2, 0) and (-3, 0)—verifies the given zeros! Choice A correctly factors and states the x-intercepts. Choice B swaps the signs, expanding to x² - x - 6—middle term sign flips; verify by plugging in zeros. Test proposed zeros in original: f(2) = 4 + 2 - 6 = 0, f(-3) = 9 - 3 - 6 = 0—perfect check; build confidence with this method!

Question 14

A parabola crosses the x-axis where g(x)=x2−25g(x)=x^2-25 equals zero. Factor g(x)g(x) and identify the zeros and x-intercepts.

  1. g(x)=(x−5)(x+5)g(x)=(x-5)(x+5); zeros: x=5,−5x=5,-5; x-intercepts: (5,0),(−5,0)(5,0),(-5,0) (correct answer)
  2. g(x)=(x−25)(x+1)g(x)=(x-25)(x+1); zeros: x=25,−1x=25,-1; x-intercepts: (25,0),(−1,0)(25,0),(-1,0)
  3. g(x)=(x−5)2g(x)=(x-5)^2; zero: x=5x=5; x-intercept: (5,0)(5,0)
  4. g(x)=(x+5)2g(x)=(x+5)^2; zero: x=−5x=-5; x-intercept: (−5,0)(-5,0)
Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x)=a(x−r)(x−s)f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x−r)=0(x - r) = 0 gives x=rx = r, and setting (x−s)=0(x - s) = 0 gives x=sx = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r,0)(r, 0) and (s,0)(s, 0) where the graph crosses the x-axis. For g(x)=x2−25g(x) = x^2 - 25, recognize it as a difference of squares: (x−5)(x+5)(x - 5)(x + 5); set to zero for x=5x = 5 or x=−5x = -5, with x-intercepts (5,0)(5, 0) and (−5,0)(-5, 0)—this symmetry is common in even-powered terms! Choice A correctly factors using the difference of squares and identifies the zeros accurately. Choice C mistakenly treats it as a perfect square without the negative, leading to (x−5)2=x2−10x+25(x - 5)^2 = x^2 - 10x + 25, which doesn't match—always verify by expanding back to the original. Remember, difference of squares is x2−a2=(x−a)(x+a)x^2 - a^2 = (x - a)(x + a); for zeros, solve each factor, and you'll see symmetric roots around zero—great for quick sketching!

Question 15

Factor the quadratic function f(x)=x2+7x+12f(x)=x^2+7x+12 and use the factored form to find the zeros (and thus the x-intercepts) of the graph.

  1. f(x)=(x−3)(x−4)f(x)=(x-3)(x-4); zeros: x=3,4x=3,4; x-intercepts: (3,0),(4,0)(3,0),(4,0)
  2. f(x)=(x+2)(x+6)f(x)=(x+2)(x+6); zeros: x=−2,−6x=-2,-6; x-intercepts: (−2,0),(−6,0)(-2,0),(-6,0)
  3. f(x)=(x+3)(x−4)f(x)=(x+3)(x-4); zeros: x=−3,4x=-3,4; x-intercepts: (−3,0),(4,0)(-3,0),(4,0)
  4. f(x)=(x+3)(x+4)f(x)=(x+3)(x+4); zeros: x=−3,−4x=-3,-4; x-intercepts: (−3,0),(−4,0)(-3,0),(-4,0) (correct answer)
Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. To factor x² + 7x + 12 and find zeros: look for two numbers that multiply to 12 and add to 7 (3 and 4), so (x + 3)(x + 4); set to zero for x = -3 or x = -4, with x-intercepts (-3, 0) and (-4, 0)—factored form makes this straightforward! Choice A correctly factors the quadratic and identifies both zeros by setting each factor equal to zero and solving. Choice B has a sign error in the factors: using negative numbers for positive coefficients leads to incorrect factors like (x - 3)(x - 4), which expands to x² - 7x + 12, flipping the middle sign—always check by expanding! Factoring trinomial x² + bx + c: find two numbers that multiply to c and add to b, then form (x + first)(x + second); for zeros, remember the sign flips, as in (x + 3) = 0 gives x = -3—practice this to master it quickly!

Question 16

Solve the equation by factoring and using the zero product property: 3x2−12=03x^2-12=0. Then list the zeros of the related function y=3x2−12y=3x^2-12 and the x-intercepts.

  1. Zeros: x=2x=2 only; x-intercept: (2,0)(2,0)
  2. Zeros: x=±4x=\pm 4; x-intercepts: (−4,0),(4,0)(-4,0),(4,0)
  3. Zeros: x=−2x=-2 only; x-intercept: (−2,0)(-2,0)
  4. Zeros: x=±2x=\pm 2; x-intercepts: (−2,0),(2,0)(-2,0),(2,0) (correct answer)
Explanation: This question tests solving equations by factoring and relating to function zeros (x-intercepts)—builds equation-solving skills. Factor 3x² - 12 = 0 → 3(x² - 4) = 0 → 3(x - 2)(x + 2) = 0. Zeros: ±2 (3≠0). X-intercepts: (-2, 0), (2, 0). You're doing great! Choice A correctly identifies zeros. Choice B mistakes the constant; -12/3 = -4, but √4=2, not 4. Factor completely! Difference of squares after factoring out common factor. Example: 2x² - 8 = 0 → 2(x² - 4)=0 → 2(x-2)(x+2)=0, zeros ±2.

Question 17

To sketch the parabola y=x2−25y=x^2-25, first factor to reveal the zeros. What are the zeros and corresponding x-intercepts?

  1. Zeros: x=25,−25x=25,-25; x-intercepts: (25,0),(−25,0)(25,0),(-25,0)
  2. Zeros: x=5,−5x=5,-5; x-intercepts: (5,0),(−5,0)(5,0),(-5,0) (correct answer)
  3. Zeros: x=5x=5 only; x-intercept: (5,0)(5,0)
  4. Zeros: x=0,25x=0,25; x-intercepts: (0,0),(25,0)(0,0),(25,0)
Explanation: This question tests your ability to recognize and factor a difference of squares, then use the factored form to find zeros and x-intercepts for sketching the parabola. Factoring a quadratic like x² - c (difference of squares) into (x - √c)(x + √c) reveals the zeros immediately via the zero product property. Setting each factor to zero gives the roots, which are symmetric around zero for this form. These zeros are the x-intercepts, key points for graphing the parabola that opens upward from the origin. Great job spotting this special case—it simplifies everything! To factor x² - 25: recognize it as (x)² - (5)² = (x - 5)(x + 5). Set to zero: x - 5 = 0 so x = 5, or x + 5 = 0 so x = -5. Zeros are x = 5 and x = -5, with x-intercepts (5, 0) and (-5, 0). This symmetry makes sense since the vertex is at x=0. Choice B correctly identifies the zeros and intercepts from the proper factoring. Choice A mistakenly uses ±25, perhaps confusing with x² - 625 = (x - 25)(x + 25), but always check by expanding back: (x - 5)(x + 5) = x² - 25, perfect match! For difference of squares a² - b² = (a - b)(a + b), zeros at x = b and x = -b. Example: x² - 16 = (x - 4)(x + 4), zeros ±4. If it's x² + c, no real factors, but here it's minus. Remember to factor out common factors first if present, like 4x² - 9 = (2x - 3)(2x + 3). Keep practicing these patterns—they speed up solving!

Question 18

To sketch f(x)=x2+2x−8f(x)=x^2+2x-8, first factor it to find the zeros. Which choice gives the correct factored form and x-intercepts?

  1. f(x)=(x+4)(x−2)f(x)=(x+4)(x-2); zeros: x=−4,2x=-4,2; x-intercepts: (−4,0),(2,0)(-4,0),(2,0) (correct answer)
  2. f(x)=(x−4)(x+2)f(x)=(x-4)(x+2); zeros: x=4,−2x=4,-2; x-intercepts: (4,0),(−2,0)(4,0),(-2,0)
  3. f(x)=(x+8)(x−1)f(x)=(x+8)(x-1); zeros: x=−8,1x=-8,1; x-intercepts: (−8,0),(1,0)(-8,0),(1,0)
  4. f(x)=(x+2)2−12f(x)=(x+2)^2-12; zeros: x=−2±12x=-2\pm 12; x-intercepts: (−14,0),(10,0)(-14,0),(10,0)
Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For f(x) = x² + 2x - 8, numbers multiply to -8, add to 2: 4 and -2, so (x + 4)(x - 2); zeros x = -4 and x = 2, x-intercepts (-4, 0) and (2, 0)—one positive, one negative! Choice A correctly factors and finds the zeros. Choice B swaps signs, expanding to x² - 2x - 8—close but middle term wrong; always FOIL to check! For mixed signs, choose pairs with opposite signs; this helps sketch parabolas quickly—keep going, you're doing great!

Question 19

A profit function is P(x)=x2−8x+15P(x)=x^2-8x+15, where zeros represent break-even points. Factor P(x)P(x) and find the break-even x-values.

  1. P(x)=(x+3)(x+5)P(x)=(x+3)(x+5); break-even x-values: x=−3,−5x=-3,-5
  2. P(x)=(x−1)(x−15)P(x)=(x-1)(x-15); break-even x-values: x=1,15x=1,15
  3. P(x)=(x−3)(x−5)P(x)=(x-3)(x-5); break-even x-values: x=3,5x=3,5 (correct answer)
  4. P(x)=(x−8)(x+7)P(x)=(x-8)(x+7); break-even x-values: x=8,−7x=8,-7
Explanation: This question tests your ability to factor quadratic expressions and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations; here, zeros are break-even points. Factoring a quadratic into form f(x) = a(x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s—these are the x-intercepts of the parabola at points (r, 0) and (s, 0) where the graph crosses the x-axis. For P(x) = x² - 8x + 15, find numbers multiplying to 15 adding to -8: -3 and -5, so (x - 3)(x - 5); break-even at x = 3 and x = 5—positive zeros make sense for profits! Choice A correctly factors and identifies the break-even points. Choice B flips signs, giving positive factors that expand to x² + 8x + 15—signs are key, opposite for negative sum! Apply this to real-world models: factors reveal intercepts directly; practice sign relationships to avoid common errors—you're building strong skills!

Question 20

Factor and find the zero(s) of h(x)=x2−6x+9h(x)=x^2-6x+9. Then state the x-intercept(s) of the graph.

  1. h(x)=(x−3)2h(x)=(x-3)^2; zero: x=3x=3 (double root); x-intercept: (3,0)(3,0) (correct answer)
  2. h(x)=(x+3)2h(x)=(x+3)^2; zero: x=−3x=-3; x-intercept: (−3,0)(-3,0)
  3. h(x)=(x−9)(x+1)h(x)=(x-9)(x+1); zeros: x=9,−1x=9,-1; x-intercepts: (9,0),(−1,0)(9,0),(-1,0)
  4. h(x)=(x−3)(x+3)h(x)=(x-3)(x+3); zeros: x=3,−3x=3,-3; x-intercepts: (3,0),(−3,0)(3,0),(-3,0)
Explanation: This question tests your ability to recognize and factor perfect square trinomials to find zeros (x-intercepts)—key for understanding quadratic behavior like touching the x-axis at a double root. Factoring into (x - r)² shows a double zero at x = r, meaning the parabola touches the x-axis at (r, 0) without crossing—use zero product property on the repeated factor. The zero is r (with multiplicity 2). This indicates the vertex is on the x-axis. Factoring reveals this structure easily! To factor x² - 6x + 9: check if it's (x - k)² = x² - 2kx + k²; here, -2k = -6 so k = 3, and k² = 9 matches. So (x - 3)². Zero: x - 3 = 0 gives x = 3 (double). X-intercept: (3, 0). You're building strong skills! Choice A correctly identifies the perfect square and the double root. Choice B has the wrong sign; (x + 3)² = x² + 6x + 9, but original has -6x, so signs matter—middle term sign determines the factor sign. Perfect square trinomial: x² ± 2kx + k² = (x ± k)². Example: x² + 10x + 25 = (x + 5)², zero -5. Verify by expanding!