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Algebra

Algebra Help: Factor Quadratics To Find Zeros

Review real example questions for Factor Quadratics To Find Zeros in Algebra.

Question 1

What values of xxx make x2+2x−8=0x^2 + 2x - 8 = 0x2+2x−8=0? (Factor first, then use the Zero Product Property.)

  1. x=−4x=-4x=−4 and x=2x=2x=2
  2. x=4x=4x=4 and x=−2x=-2x=−2
  3. x=−2x=-2x=−2 and x=4x=4x=4
  4. x=−8x=-8x=−8 and x=1x=1x=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

To graph y=x2+7x+12y=x^2+7x+12y=x2+7x+12, first factor to find where the parabola crosses the xxx-axis. Where does the graph cross the xxx-axis?

  1. (3,0)(3,0)(3,0) and (4,0)(4,0)(4,0)
  2. (0,−3)(0,-3)(0,−3) and (0,−4)(0,-4)(0,−4)
  3. (−3,0)(-3,0)(−3,0) and (−4,0)(-4,0)(−4,0)
  4. (−2,0)(-2,0)(−2,0) and (−6,0)(-6,0)(−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

A company’s break-even points occur when its profit is zero. If P(x)=−x2+10x−16P(x)=-x^2+10x-16P(x)=−x2+10x−16, for what values of xxx is P(x)=0P(x)=0P(x)=0?

  1. x=4x=4x=4 and x=4x=4x=4
  2. x=1x=1x=1 and x=16x=16x=16
  3. x=2x=2x=2 and x=8x=8x=8
  4. x=−2x=-2x=−2 and x=−8x=-8x=−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 x2+bx+cx^2 + bx + cx2+bx+c, we need to find two numbers that multiply to give ccc (the constant) and add to give bbb (the coefficient of x), then write the factored form as (x+[first number])(x+[second number])(x + \text{[first number]})(x + \text{[second number]})(x+[first number])(x+[second number]). Let's factor P(x)=−x2+10x−16P(x) = -x^2 + 10x - 16P(x)=−x2+10x−16: First, factor out −1-1−1 to get −(x2−10x+16)-(x^2 - 10x + 16)−(x2−10x+16). Now we need two numbers that multiply to 161616 and add to −10-10−10. Those numbers are −2-2−2 and −8-8−8 because (−2)⋅(−8)=16(-2) \cdot (-8) = 16(−2)⋅(−8)=16 and (−2)+(−8)=−10(-2) + (-8) = -10(−2)+(−8)=−10. So the factored form is −(x−2)(x−8)-(x - 2)(x - 8)−(x−2)(x−8). Now we use the Zero Product Property: set each factor to zero and solve. x−2=0x - 2 = 0x−2=0 gives x=2x = 2x=2, and x−8=0x - 8 = 0x−8=0 gives x=8x = 8x=8. Those are the zeros! Choice A is correct because it properly identifies the zeros as x=2x = 2x=2 and x=8x = 8x=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)(x - p)(x - q)(x−p)(x−q), the zeros are x=px = px=p and x=qx = qx=q. Notice how the signs flip! If you have (x−3)(x - 3)(x−3), the zero is x=3x = 3x=3 (positive), and if you have (x+5)(x + 5)(x+5), the zero is x=−5x = -5x=−5 (negative). The sign in the factor is opposite the sign of the zero.

Question 4

The zeros of f(x)=x2−4x−5f(x) = x^2 - 4x - 5f(x)=x2−4x−5 are needed to find the xxx-intercepts of its graph. What are the zeros?

  1. x=0x = 0x=0 and x=−5x = -5x=−5
  2. x=−1x = -1x=−1 and x=5x = 5x=5
  3. x=1x = 1x=1 and x=−5x = -5x=−5
  4. x=−5x = -5x=−5 and x=−1x = -1x=−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

Factor x2−9x^2-9x2−9 and identify the zeros.

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

The zeros of f(x)=x2+x−12f(x)=x^2+x-12f(x)=x2+x−12 are claimed to be x=3x=3x=3 and x=−4x=-4x=−4. Verify by factoring: what are the zeros of f(x)f(x)f(x)?

  1. x=3x=3x=3 and x=−4x=-4x=−4
  2. x=3x=3x=3 and x=4x=4x=4
  3. x=−3x=-3x=−3 and x=−4x=-4x=−4
  4. x=−3x=-3x=−3 and x=4x=4x=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

Factor 2x2+7x+32x^2 + 7x + 32x2+7x+3 and use the factors to find the zeros.

  1. Zeros: x=−3x=-3x=−3 and x=−12x=-\tfrac{1}{2}x=−21​
  2. Zeros: x=3x=3x=3 and x=12x=\tfrac{1}{2}x=21​
  3. Zeros: x=−1x=-1x=−1 and x=−3x=-3x=−3
  4. Zeros: x=−32x=-\tfrac{3}{2}x=−23​ and x=−1x=-1x=−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

The revenue function for selling xxx items is R(x)=x2−6x+5R(x)=x^2-6x+5R(x)=x2−6x+5. Break-even occurs when R(x)=0R(x)=0R(x)=0. What are the solutions to x2−6x+5=0x^2-6x+5=0x2−6x+5=0?

  1. x=1x=1x=1 and x=5x=5x=5
  2. x=−1x=-1x=−1 and x=−5x=-5x=−5
  3. x=2x=2x=2 and x=3x=3x=3
  4. x=0x=0x=0 and x=5x=5x=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

Factor x2+6x+9x^2 + 6x + 9x2+6x+9 and find its zeros.

  1. Zeros: x=0x = 0x=0 and x=−9x = -9x=−9
  2. Zeros: x=−9x = -9x=−9 and x=1x = 1x=1
  3. Zero: x=−3x = -3x=−3
  4. Zeros: x=−3x = -3x=−3 and x=3x = 3x=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

A quadratic function has the form f(x)=x2+bx+cf(x) = x^2 + bx + cf(x)=x2+bx+c where bbb and ccc are integers. If the function can be factored as (x+3)(x−5)(x + 3)(x - 5)(x+3)(x−5), what is the value of b−cb - cb−c?

  1. -17
  2. 17
  3. 13
  4. -13
Explanation: Expand (x+3)(x−5)(x + 3)(x - 5)(x+3)(x−5): $(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15.Comparingwith. Comparing with .Comparingwithf(x) = x^2 + bx + c,wehave, we have ,wehaveb = -2andandandc = -15.Therefore. Therefore .Thereforeb - c = (-2) - (-15) = -2 + 15 = 13.ChoiceAresultsfromcalculating. Choice A results from calculating .ChoiceAresultsfromcalculatingc - binsteadofinstead ofinsteadofb - c.ChoiceBcomesfromsignerrorsintheexpansion.ChoiceDresultsfromincorrectlycomputing. Choice B comes from sign errors in the expansion. Choice D results from incorrectly computing .ChoiceBcomesfromsignerrorsintheexpansion.ChoiceDresultsfromincorrectlycomputingb - casasas-2 - 15 = -17$$ and then making a sign error.