Factoring Polynomials Worksheet Answer Key – Algebra 1

Problem 1 Take Out the GCF

The GCF of 12 and 18 is 6, and the lowest power of x is x². Divide each term: 12x³ ÷ 6x² = 2x and 18x² ÷ 6x² = 3.

6x²(2x + 3)

Problem 2 GCF with Two Variables

GCF: 5, a and b, so 5ab. Divide: 15a²b ÷ 5ab = 3a and 20ab² ÷ 5ab = 4b.

5ab(3a − 4b)

Problem 3 Positive Trinomial

Need product 20 and sum 9: 4 and 5.

Check: x² + 5x + 4x + 20 ✓.

(x + 4)(x + 5)

Problem 4 Trinomial with Negative Middle

Product +24 and sum −11 means both numbers are negative: −3 and −8.

(x − 3)(x − 8)

Problem 5 Negative Constant, Larger Positive

Product −35 means the signs differ. Sum +2: 7 and −5.

(x + 7)(x − 5)

Problem 6 Negative Constant, Larger Negative

Product −54, sum −3: −9 and 6.

Check: x² + 6x − 9x − 54 = x² − 3x − 54 ✓.

(x − 9)(x + 6)

Problem 7 Difference of Squares

49m² = (7m)² and 100 = 10².

(7m + 10)(7m − 10)

Problem 8 Perfect Square Trinomials

1. x² and 8²; middle term 2(x)(8) = 16x ✓. So (x + 8)².
2. (3y)² and 5²; middle term −2(3y)(5) = −30y ✓. So (3y − 5)².

(a) (x + 8)²; (b) (3y − 5)²

Problem 9 GCF Then Trinomial

Take out 3: 3(x² − 6x + 8). Then −2 and −4 give product 8 and sum −6.

3(x − 2)(x − 4)

Problem 10 Leading Coefficient 2

ac = 24; the numbers 3 and 8 have sum 11. Split: 2x² + 3x + 8x + 12.

Group: x(2x + 3) + 4(2x + 3).

(2x + 3)(x + 4)

Problem 11 Leading Coefficient 3

ac = 24 and the sum is −10, so use −6 and −4. Split: 3x² − 6x − 4x + 8.

Group: 3x(x − 2) − 4(x − 2).

(3x − 4)(x − 2)

Problem 12 Mixed Signs with a = 6

ac = −90 and the sum is 1, so use 10 and −9. Split: 6x² + 10x − 9x − 15.

Group: 2x(3x + 5) − 3(3x + 5).

(2x − 3)(3x + 5)

Problem 13 Grouping Four Terms

Group: (x³ + 5x²) + (2x + 10) = x²(x + 5) + 2(x + 5).

(x + 5)(x² + 2)

Problem 14 Grouping with Negatives

Group: 3x²(2x − 3) + 2(2x − 3).

(2x − 3)(3x² + 2)

Problem 15 Solve by Factoring

Factor: (x − 8)(x + 3) = 0. Set each factor equal to zero: x − 8 = 0 or x + 3 = 0.

x = 8 or x = −3

Problem 16 Zeros of a Quadratic Function

ac = −30 and the sum is −7, so use −10 and 3. Split: 2x² − 10x + 3x − 15 = 2x(x − 5) + 3(x − 5) = (2x + 3)(x − 5).

2x + 3 = 0 gives x = −32; x − 5 = 0 gives x = 5.

x = −32 and x = 5

Problem 17 Missing Side of a Rectangle

Factor: x² + 13x + 42 = (x + 6)(x + 7), so the length is x + 7.

For x = 5: width 11 feet and length 12 feet. Area 11 × 12 = 132, and 25 + 65 + 42 = 132 ✓.

Length x + 7; 12 ft by 11 ft

Problem 18 Ball in the Air

Take out −16t: −16t(t − 4).

Set equal to 0: t = 0 (thrown) or t = 4 (lands). The graph meets the ground at 0 and 4 seconds.

−16t(t − 4); at ground level at 0 s and 4 s

Problem 19 Factor Completely, Two Rounds

Take out 2: 2(x⁴ − 16). Then x⁴ − 16 = (x²)² − 4² = (x² + 4)(x² − 4).

Factor again: x² − 4 = (x + 2)(x − 2). The factor x² + 4 stays.

2(x² + 4)(x + 2)(x − 2)

Problem 20 See the Structure

Treat x + 3 as one quantity: u² − 5² = (u + 5)(u − 5) with u = x + 3.

(x + 3 + 5)(x + 3 − 5) = (x + 8)(x − 2).

Expand: (x + 3)² − 25 = x² + 6x − 16 and (x + 8)(x − 2) = x² + 6x − 16 ✓.

(x + 8)(x − 2)