Factor Quadratics

Grade 9 · algebra · 97 practice problems · read aloud

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Factoring Quadratics: Unlocking the Roots 🔑

Factoring is like "un-multiplying" a quadratic expression. It rewrites ax² + bx + c as a product of two simpler expressions. This is super useful because it helps us find the roots (or x-intercepts) of a parabola, which is essential for solving equations and graphing.

Step-by-Step Guide: The AC Method

  1. Identify a, b, and c: From the standard form ax² + bx + c.
  2. Multiply a and c: Find the product a*c.
  3. Find two magic numbers: Look for two numbers that multiply to a*c and add to b.
  4. Rewrite the middle term: Split bx into two terms using your magic numbers.
  5. Factor by grouping: Group the first two and last two terms, and factor out the Greatest Common Factor (GCF) from each.
  6. Final factored form: You should now have a common binomial factor. Write your answer as (binomial)(binomial).

Visual Examples

Example 1: Factor x² + 5x + 6

  1. a=1, b=5, c=6
  2. a*c = 6
  3. Magic numbers: 2 and 3 (2*3=6, 2+3=5) ✅
  4. Rewrite: x² + 2x + 3x + 6
  5. Group: (x² + 2x) + (3x + 6) → x(x + 2) + 3(x + 2)
  6. Final: (x + 2)(x + 3)

Example 2: Factor 2x² - 7x - 15

  1. a=2, b=-7, c=-15
  2. a*c = -30
  3. Magic numbers: -10 and 3 (-10*3=-30, -10+3=-7) ✅
  4. Rewrite: 2x² - 10x + 3x - 15
  5. Group: (2x² - 10x) + (3x - 15) → 2x(x - 5) + 3(x - 5)
  6. Final: (x - 5)(2x + 3)

Common Mistakes to Avoid 🚫

  • Forgetting the GCF first: Always check if all terms have a common factor to pull out before starting the AC method.
  • Incorrect signs: Be extra careful with negative values for b and c. The magic numbers must multiply to a*c and add to b.
  • Rushing the final factor: After grouping, make sure the binomial in the parentheses is exactly the same to factor it out completely.

Tips & Tricks

  • M.A.P. (Multiply, Add, Pair): A great memory aid. Multiply a*c, Add to b, Pair for grouping.
  • Spot the Easy Ones: If a=1, you just need two numbers that multiply to c and add to b.
  • Check Your Work! You can always FOIL your final answer to see if you get the original quadratic back.

How to Practice

Start simple and build up complexity!

  1. Begin with quadratics where a=1 (e.g., x² + 6x + 8).
  2. Move to quadratics where a>1 (e.g., 2x² + 9x + 4).
  3. Challenge yourself with negative terms (e.g., 3x² - x - 10).
  4. Try creating your own problems and factoring them, or swap with a friend.

Practice problems

6 of the 97, worked through step by step — try them before opening the answer.

1 x² + 5x + 6 = ?

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term.

Show the answer

Answer: (x + 2)(x + 3)

  1. Look for two numbers that multiply to the constant term (6) and add up to the coefficient of the middle term (5).
  2. List pairs of factors of 6: - 1 and 6: 1 + 6 = 7 (not 5) - 2 and 3: 2 + 3 = 5 (yes, this works) - Also -1 and -6, -2 and -3, but these give negative sums, so we don't need them since our middle term is positive.
  3. Since 2 and 3 multiply to 6 and add to 5, we can write the factored form as: (x + 2)(x + 3)
  4. Check by expanding (x + 2)(x + 3): - First: x * x = x^2 - Outer: x * 3 = 3x - Inner: 2 * x = 2x - Last: 2 * 3 = 6 Combine like terms: x^2 + 3x + 2x + 6 = x^2 + 5x + 6 This matches the original expression, so the factoring is correct. Final answer: (x + 2)(x + 3)

Let's factor the quadratic expression: x^2 + 5x + 6.

2 x² - 5x + 6 = ?

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term. For example, in x² + 7x + 12, the numbers 3 and 4 work because 3×4=12 and 3+4=7.

Show the answer

Answer: (x-2)(x-3)

  1. Understand the factoring pattern. When we expand (x + a)(x + b), we get: x^2 + (a + b)x + a*b. So for x^2 - 5x + 6, we need: a + b = -5 a * b = 6
  2. Look for two numbers whose product is 6 and whose sum is -5. Since the product is positive (6), a and b must have the same sign. Since the sum is negative (-5), a and b must both be negative.
  3. List pairs of negative integers that multiply to 6. Possible pairs: (-1, -6): sum = -7, product = 6 (-2, -3): sum = -5, product = 6
  4. Identify the correct pair. We see that -2 and -3 satisfy: -2 + (-3) = -5 (-2)*(-3) = 6
  5. Write the factored form. So a = -2, b = -3. Thus: x^2 - 5x + 6 = (x - 2)(x - 3)
  6. Check by expanding (x - 2)(x - 3): x*x = x^2 x*(-3) = -3x (-2)*x = -2x (-2)*(-3) = 6 Combine: x^2 - 3x - 2x + 6 = x^2 - 5x + 6. This matches the original expression. Final answer: (x - 2)(x - 3)

We are given the quadratic expression: x^2 - 5x + 6. Our goal is to factor it into the form (x + a)(x + b).

3 2x² + 7x + 6 = ?

Hint: Look for two numbers that multiply to the product of the leading coefficient and constant term, and add to the middle coefficient

Show the answer

Answer: (2x + 3)(x + 2)

When factoring quadratic expressions of the form ax² + bx + c, we use the AC method where we find two numbers that multiply to a×c and add to b. Then we rewrite the middle term using these numbers and factor by grouping.

4 2x² + 7x + 3 = ?

Hint: Look for two numbers that multiply to give the product of the leading coefficient and the constant term, and add to give the middle coefficient. For example, in ax² + bx + c, find numbers that multiply to a×c and add to b.

Show the answer

Answer: (2x + 1)(x + 3)

  1. Identify coefficients: a = 2, b = 7, c = 3
  2. Multiply a × c = 2 × 3 = 6
  3. Find two numbers that multiply to 6 and add to 7: 6 and 1
  4. Rewrite the middle term using these numbers: 2x² + 6x + 1x + 3
  5. Factor by grouping: (2x² + 6x) + (1x + 3) = 2x(x + 3) + 1(x + 3)
  6. Factor out the common binomial: (2x + 1)(x + 3)

The answer is (2x + 1)(x + 3).

5 x² - 9x + 20 = ?

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term.

Show the answer

Answer: (x-4)(x-5)

  1. Identify coefficients: a=1, b=-9, c=20
  2. Find two numbers that multiply to 20 and add to -9
  3. The numbers are -4 and -5 because (-4)×(-5)=20 and (-4)+(-5)=-9
  4. Write the factored form: (x-4)(x-5)

The answer is (x-4)(x-5).

6 x² + 8x + 15 = ?

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the middle term

Show the answer

Answer: (x+3)(x+5)

  1. Identify the quadratic expression: x² + 8x + 15
  2. Find two numbers that multiply to 15 (the constant term) and add to 8 (the coefficient of x)
  3. The numbers 3 and 5 satisfy these conditions because 3 × 5 = 15 and 3 + 5 = 8
  4. Write the factored form using these numbers: (x + 3)(x + 5)
  5. Verify by expanding: (x + 3)(x + 5) = x² + 5x + 3x + 15 = x² + 8x + 15

The answer is (x+3)(x+5).

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