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ACT Math

ACT Math Help: Quadratics And Polynomials

Review real example questions for Quadratics And Polynomials in ACT Math.

Question 1

Which is the correct factorization of x2−6x+8x^2 - 6x + 8x2−6x+8?

  1. (x+3)(x+3)(x + 3)(x + 3)(x+3)(x+3)
  2. (x+4)(x+2)(x + 4)(x + 2)(x+4)(x+2)
  3. (x−3)(x−3)(x - 3)(x - 3)(x−3)(x−3)
  4. (x−4)(x−2)(x - 4)(x - 2)(x−4)(x−2)
Explanation: To factor x² - 6x + 8, we need two numbers that multiply to 8 and add to -6. The numbers are -4 and -2 since (-4)(-2) = 8 and -4 + (-2) = -6. Therefore, x² - 6x + 8 = (x - 4)(x - 2). Choice C represents a perfect square that would give x² - 6x + 9.

Question 2

The polynomial p(x)=x3−4x2−7x+10p(x) = x^3 - 4x^2 - 7x + 10p(x)=x3−4x2−7x+10 has (x−1)(x - 1)(x−1) as a factor. What are all the other zeros of p(x)p(x)p(x)?

  1. −5-5−5 and 222
  2. 555 and −2-2−2
  3. 555 and 222
  4. −5-5−5 and −2-2−2
Explanation: This is a polynomial factors and zeros question testing synthetic division and factoring. Choice B (5 and −2) is correct — since (x − 1) is a factor, divide p(x) by (x − 1) using synthetic division: coefficients 1, −4, −7, 10 divided by root 1: bring down 1; 1×1 − 4 = −3; −3×1 − 7 = −10; −10×1 + 10 = 0 ✓. Quotient: x² − 3x − 10 = (x − 5)(x + 2). Zeros: x = 5 and x = −2. Choice A (−5 and 2) flips both signs of the correct zeros. Choice C (5 and 2) gets x = 5 correct but uses x = 2 instead of x = −2, likely from factoring x² − 3x − 10 as (x − 5)(x − 2) (wrong sign on the second factor). Choice D (−5 and −2) negates both correct zeros. Pro tip: After dividing out the known factor, you're left with a simpler polynomial to factor. Here: (x − 1)(x² − 3x − 10) = 0. Factor x² − 3x − 10: find two numbers that multiply to −10 and sum to −3: −5 and +2. So (x − 5)(x + 2) = 0 → x = 5 or x = −2.

Question 3

Which of the following is equivalent to (5x2−3x+7)−(2x2+4x−1)(5x^2 - 3x + 7) - (2x^2 + 4x - 1)(5x2−3x+7)−(2x2+4x−1)?

  1. 3x2−7x+83x^2 - 7x + 83x2−7x+8
  2. 3x2+x+63x^2 + x + 63x2+x+6
  3. 7x2+x+67x^2 + x + 67x2+x+6
  4. 7x2−7x+87x^2 - 7x + 87x2−7x+8
Explanation: This is a polynomial operations question testing distribution of a negative sign. Choice A (3x² − 7x + 8) is correct — distribute the negative sign: (5x² − 3x + 7) − (2x² + 4x − 1) = 5x² − 3x + 7 − 2x² − 4x + 1. Key step: −(−1) = +1. Combine: (5−2)x² + (−3−4)x + (7+1) = 3x² − 7x + 8. Choice B (3x² + x + 6) correctly subtracts x² terms but fails to distribute the negative on the x term: −3x − (4x) treated as −3x + 4x = +x, and −(−1) = −1 instead of +1. Choice C (7x² + x + 6) adds x² terms instead of subtracting (5 + 2 = 7) and makes both sign errors. Choice D (7x² − 7x + 8) adds x² terms (5 + 2 = 7) but correctly handles x and constant terms. Pro tip: When subtracting a polynomial, distribute the negative sign to EVERY term inside the parentheses. Rewrite −(2x² + 4x − 1) as (−2x² − 4x + 1) before combining. The term that trips most students is −(−1) = +1.

Question 4

What is the value of f(3)f(3)f(3) for f(x)=x2−4x+4f(x) = x^2 - 4x + 4f(x)=x2−4x+4?

  1. 1
  2. 0
  3. 4
  4. 3
Explanation: To find f(3)f(3)f(3) for f(x)=x2−4x+4f(x) = x^2 - 4x + 4f(x)=x2−4x+4, we substitute x=3x = 3x=3 into the function. We get f(3)=(3)2−4(3)+4=9−12+4=1f(3) = (3)^2 - 4(3) + 4 = 9 - 12 + 4 = 1f(3)=(3)2−4(3)+4=9−12+4=1. Notice that this function can also be written as (x−2)2(x - 2)^2(x−2)2, so f(3)=(3−2)2=12=1f(3) = (3 - 2)^2 = 1^2 = 1f(3)=(3−2)2=12=1.

Question 5

What is (x−1)(x+4)(x - 1)(x + 4)(x−1)(x+4)?

  1. x2+3x−4x^2 + 3x - 4x2+3x−4
  2. x2+4x−1x^2 + 4x - 1x2+4x−1
  3. x2+3x+4x^2 + 3x + 4x2+3x+4
  4. x2−4x+1x^2 - 4x + 1x2−4x+1
Explanation: To multiply (x−1)(x+4)(x - 1)(x + 4)(x−1)(x+4) using FOIL: First terms: x⋅x=x2x \cdot x = x^2x⋅x=x2; Outer terms: x⋅4=4xx \cdot 4 = 4xx⋅4=4x; Inner terms: (−1)⋅x=−x(-1) \cdot x = -x(−1)⋅x=−x; Last terms: (−1)⋅4=−4(-1) \cdot 4 = -4(−1)⋅4=−4. Combining these gives x2+4x−x−4=x2+3x−4x^2 + 4x - x - 4 = x^2 + 3x - 4x2+4x−x−4=x2+3x−4.

Question 6

Factor: x2−9x^2 - 9x2−9

  1. (x−3)(x−3)(x - 3)(x - 3)(x−3)(x−3)
  2. (x−9)(x+1)(x - 9)(x + 1)(x−9)(x+1)
  3. (x−3)(x+3)(x - 3)(x + 3)(x−3)(x+3)
  4. (x+9)(x−1)(x + 9)(x - 1)(x+9)(x−1)
Explanation: To factor x2−9x^2 - 9x2−9, we recognize this as a difference of squares pattern a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)a2−b2=(a+b)(a−b). Here we have x2−32x^2 - 3^2x2−32, so the factorization is (x+3)(x−3)(x + 3)(x - 3)(x+3)(x−3). We can verify: (x+3)(x−3)=x2−3x+3x−9=x2−9(x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9(x+3)(x−3)=x2−3x+3x−9=x2−9.

Question 7

What are the solutions to the quadratic equation x2−4x+4=0x^2 - 4x + 4 = 0x2−4x+4=0?

  1. -2 and -2
  2. -2 and 2
  3. 2 and 2
  4. 4 and -4
Explanation: To solve x² - 4x + 4 = 0, we recognize this as a perfect square trinomial. The equation factors as (x - 2)² = 0, which means x - 2 = 0, so x = 2. Since this is a repeated root, both solutions are x = 2. The solutions are 2 and 2.

Question 8

What is the value of f(0)f(0)f(0) for f(x)=3x2−4x+7f(x) = 3x^2 - 4x + 7f(x)=3x2−4x+7?

  1. 7
  2. 3
  3. 0
  4. 4
Explanation: To find f(0)f(0)f(0) for f(x)=3x2−4x+7f(x) = 3x^2 - 4x + 7f(x)=3x2−4x+7, we substitute x=0x = 0x=0 into the function. We get f(0)=3(0)2−4(0)+7=0−0+7=7f(0) = 3(0)^2 - 4(0) + 7 = 0 - 0 + 7 = 7f(0)=3(0)2−4(0)+7=0−0+7=7. When evaluating polynomial functions at zero, all terms with xxx become zero, leaving only the constant term.

Question 9

A polynomial used in a calculation is x2−6x+9x^2-6x+9x2−6x+9. Factor: x2−6x+9x^2-6x+9x2−6x+9.

  1. (x−3)(x+3)\left(x-3\right)\left(x+3\right)(x−3)(x+3)
  2. (x−3)2\left(x-3\right)^2(x−3)2
  3. (x+3)2\left(x+3\right)^2(x+3)2
  4. (x−9)(x+1)\left(x-9\right)\left(x+1\right)(x−9)(x+1)
Explanation: The expression x² - 6x + 9 is a perfect square trinomial of the form a² - 2ab + b² where a = x and b = 3. Since (-6x) = 2(x)(-3) and 9 = (-3)², we get x² - 6x + 9 = (x - 3)². We can verify: (x - 3)² = x² - 6x + 9. Choice A would be a difference of squares.

Question 10

A revenue expression is x2+x−12x^2+x-12x2+x−12. Factor: x2+x−12x^2+x-12x2+x−12.

  1. (x+3)(x−4)\left(x+3\right)\left(x-4\right)(x+3)(x−4)
  2. (x−3)(x+4)\left(x-3\right)\left(x+4\right)(x−3)(x+4)
  3. (x+2)(x−6)\left(x+2\right)\left(x-6\right)(x+2)(x−6)
  4. (x+1)(x−12)\left(x+1\right)\left(x-12\right)(x+1)(x−12)
Explanation: To factor x² + x - 12, find two numbers that multiply to -12 and add to 1. Since (4)(-3) = -12 and 4 + (-3) = 1, the factorization is (x + 4)(x - 3). Rearranging gives (x - 3)(x + 4). We can verify: (x - 3)(x + 4) = x² + 4x - 3x - 12 = x² + x - 12.