Packets  /  Factoring Quadratics
Grades 8-12Algebra

Factoring Quadratics

Master every factoring technique — GCF, trinomials, difference of squares, perfect squares, grouping — then solve quadratics by factoring. Learn the concept, practice with instant feedback, and beat the 60-second challenge.

1Greatest Common Factor (GCF)

Key idea: Before trying anything fancy, always look for a greatest common factor first — the biggest expression that divides every term. Factoring it out makes everything else easier.
  1. Find the GCF of the coefficients (the biggest number dividing all of them).
  2. Take the smallest power of x that appears in every term.
  3. Divide each term by the GCF and write: GCF(what is left).
  4. Check by distributing — you must get the original back.
Worked example
6x2 + 9x
GCF of 6 and 9 is 3; smallest power of x is x. GCF = 3x, so 3x(2x + 3). Check: 3x(2x) + 3x(3) = 6x2 + 9x.
Worked example
4x2 – 8x + 12
GCF of 4, 8, 12 is 4 (no common x). So 4(x2 – 2x + 3). Check: 4(x2) – 4(2x) + 4(3) = 4x2 – 8x + 12.
Watch out: The #1 mistake is stopping too early: 6x^2 + 9x = 3(2x^2 + 3x) is NOT finished — you can still factor out an x. Always factor completely!

2Factoring x^2 + bx + c

Key idea: Find two numbers that multiply to c and add to b. Then write (x + m)(x + n).
  1. List factor pairs of c (the constant).
  2. Find the pair that adds to b (the middle coefficient).
  3. Write (x + first)(x + second).
  4. Check with FOIL.
Worked example
x2 + 7x + 12
Need two numbers with product 12 and sum 7: 3 and 4. So (x + 3)(x + 4). Check: x2 + 4x + 3x + 12 = x2 + 7x + 12.
Worked example
x2 – 5x + 6
Need product 6, sum -5: -2 and -3. So (x – 2)(x – 3). Check: x2 – 3x – 2x + 6 = x2 – 5x + 6.
Watch out: Signs matter! If c is positive, both numbers have the same sign (match the sign of b). If c is negative, the numbers have opposite signs.

3Factoring ax^2 + bx + c (a not 1)

Key idea: Use the ac-method: multiply a and c, find two numbers with that product and sum b, then split the middle term and factor by grouping.
  1. Compute a times c.
  2. Find two numbers with product ac and sum b.
  3. Split the middle term using those two numbers.
  4. Factor by grouping.
Worked example
2x2 + 7x + 3
ac = 6. Need product 6, sum 7: 1 and 6. Split: 2x2 + x + 6x + 3. Group: x(2x + 1) + 3(2x + 1) = (x + 3)(2x + 1).
Worked example
3x2 – 10x + 8
ac = 24. Need product 24, sum -10: -4 and -6. Split: 3x2 – 4x – 6x + 8. Group: x(3x – 4) – 2(3x – 4) = (x – 2)(3x – 4).
Watch out: Always check for a GCF first! Factoring 2x^2 + 4x + 2 as 2(x^2 + 2x + 1) = 2(x + 1)^2 is much easier than the ac-method on the original.

4Difference of Squares

Key idea: a^2 – b^2 = (a – b)(a + b). Look for two perfect squares being subtracted.
  1. Check: are both terms perfect squares?
  2. Check: is there a minus sign between them?
  3. Take the square root of each term.
  4. Write (root1 – root2)(root1 + root2).
Worked example
x2 – 25
Both are perfect squares: x2 and 25 = 52. So (x – 5)(x + 5).
Worked example
4x2 – 9
4x2 = (2x)2 and 9 = 32. So (2x – 3)(2x + 3).
Watch out: Sum of squares (a^2 + b^2) does NOT factor over the reals. Only the difference works!

5Perfect Square Trinomials

Key idea: a^2 + 2ab + b^2 = (a + b)^2 and a^2 – 2ab + b^2 = (a – b)^2. The middle term is twice the product of the roots.
  1. Check the first and last terms are perfect squares.
  2. Check the middle term equals 2 times (root1)(root2).
  3. Write (root1 + root2)2 or (root1 – root2)2 matching the middle sign.
Worked example
x2 + 6x + 9
x2 = (x)2, 9 = 32, and 2(x)(3) = 6x. So (x + 3)2.
Worked example
4x2 – 12x + 9
4x2 = (2x)2, 9 = 32, and 2(2x)(3) = 12x. So (2x – 3)2.
Watch out: x^2 + 6x + 9 = (x + 3)^2, NOT (x + 3)(x + 3) expanded wrong. They are the same thing — but do not write x^2 + 9!

6Factoring by Grouping

Key idea: For four-term polynomials, group pairs, factor each pair, then factor out the common binomial.
  1. Group the terms into two pairs.
  2. Factor the GCF out of each pair.
  3. If the same binomial appears in both, factor it out.
  4. Check by multiplying.
Worked example
x3 + 2x2 + 3x + 6
Group: (x3 + 2x2) + (3x + 6) = x2(x + 2) + 3(x + 2) = (x2 + 3)(x + 2).
Worked example
2x3 – 3x2 + 4x – 6
Group: (2x3 – 3x2) + (4x – 6) = x2(2x – 3) + 2(2x – 3) = (x2 + 2)(2x – 3).
Watch out: If the binomials do not match after factoring each pair, try a different grouping — or the polynomial may not factor by grouping.

7Choosing the Strategy

Key idea: Use this order every time: (1) GCF, (2) count the terms — 2 terms: difference of squares; 3 terms: trinomial methods; 4 terms: grouping.
  1. Step 1: Factor out the GCF if there is one.
  2. Step 2: Count remaining terms.
  3. Two terms: check for difference of squares (or sum/difference of cubes).
  4. Three terms: try x2 + bx + c, then ac-method, then perfect square check.
  5. Four terms: factor by grouping.
Worked example
2x2 – 8
GCF = 2: 2(x2 – 4). Then difference of squares: 2(x – 2)(x + 2).
Worked example
x2 + 4x + 4
No GCF. Three terms, perfect square: (x + 2)2.
Watch out: Skipping the GCF step is the most common reason students get stuck — always start there!

8Solving Quadratics by Factoring

Key idea: Factor, then use the zero product property: if AB = 0, then A = 0 or B = 0.
  1. Move everything to one side so the equation equals 0.
  2. Factor completely.
  3. Set each factor equal to 0 and solve.
  4. Check each solution in the original equation.
Worked example
x2 + 5x + 6 = 0
Factor: (x + 2)(x + 3) = 0. So x + 2 = 0 or x + 3 = 0. Solutions: x = -2, x = -3.
Worked example
x2 = 9
Move to one side: x2 – 9 = 0. Factor: (x – 3)(x + 3) = 0. Solutions: x = 3, x = -3.
Watch out: You can only use the zero product property when one side is 0! Never divide both sides by x — you will lose a solution.
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60-Second Challenge

How many factoring quadratics problems can you solve in 60 seconds?