Algebra 1
Factoring
Fresh, parallel-form problems with full worked solutions — more reps for the skills in this unit, kept separate from the textbook's own problems.
Read Unit 11 in the textbook →
Lesson 11.1: Greatest common factor (GCF)
Factor out the GCF: 7x + 21. Then multiply your answer back to prove it.
Worked solutionTry it first, then open.
Numbers only. The biggest number dividing both 7 and 21 is 7; there is no x in the second term, so no variable comes out. GCF = 7. Divide each term by 7: 7x ÷ 7 = x and 21 ÷ 7 = 3.
7x + 21 = 7(x + 3)
Check (multiply back): 7·x + 7·3 = 7x + 21.
Answer: 7*(x+3)
Factor out the GCF: 12x² − 8x. Watch the sign on the second term.
Worked solutionTry it first, then open.
Two parts. Numbers: gcd(12, 8) = 4. Variables: both terms have at least one x (x² and x¹), so the lowest power x¹ comes out. GCF = 4x.
Divide each term by 4x: 12x² ÷ 4x = 3x, and −8x ÷ 4x = −2 (the minus stays).
12x² − 8x = 4x(3x − 2)
Check: 4x·3x − 4x·2 = 12x² − 8x.
Note: 2x(6x − 4) and 4(3x² − 2x) are common factors but not the greatest — the inside still shares something.
Answer: 4*x*(3*x-2)
Factor out the GCF: 20x³ + 15x². Take the lowest power of x.
Worked solutionTry it first, then open.
Numbers: gcd(20, 15) = 5. Variables: the powers present are x³ and x²; the lowest is x², so x² comes out (you can't pull x³ from the x² term). GCF = 5x².
Divide: 20x³ ÷ 5x² = 4x, and 15x² ÷ 5x² = 3.
20x³ + 15x² = 5x²(4x + 3)
Check: 5x²·4x + 5x²·3 = 20x³ + 15x².
Answer: 5*x**2*(4*x+3)
Factor out the GCF completely: 24x⁴ − 36x². (Light stretch — the inside still has an x², and that's fine.)
Worked solutionTry it first, then open.
Numbers: gcd(24, 36) = 12. Variables: powers x⁴ and x²; lowest is x². GCF = 12x².
Divide: 24x⁴ ÷ 12x² = 2x², and −36x² ÷ 12x² = −3.
24x⁴ − 36x² = 12x²(2x² − 3)
Check: 12x²·2x² − 12x²·3 = 24x⁴ − 36x².
Is the inside done? 2x² − 3 has no common factor and is not a difference of squares over the integers (3 isn't a perfect square and 2 isn't either), so 12x²(2x² − 3) is fully factored. A near-miss to avoid: pulling only 6x² gives 6x²(4x² − 6), but the inside still shares a 2 — not the greatest.
Answer: 12*x**2*(2*x**2-3)
Lesson 11.2: Factoring trinomials x²+bx+c
Factor: x² + 8x + 12.
Worked solutionTry it first, then open.
Need two numbers that multiply to c = 12 and add to b = 8. Since c > 0 the two numbers share a sign, and b > 0 forces both positive. Pairs of 12: 1·12 (sum 13), 2·6 (sum 8), 3·4 (sum 7). The pair is 2 and 6.
x² + 8x + 12 = (x + 2)(x + 6)
Check: x² + 6x + 2x + 12 = x² + 8x + 12.
Answer: (x+2)*(x+6)
Factor: x² − 9x + 20. Reason about the signs first.
Worked solutionTry it first, then open.
Two numbers multiply to c = 20 and add to b = −9. c > 0 ⇒ same sign; b < 0 ⇒ both negative. Negative pairs of 20: (−1)(−20) sum −21, (−2)(−10) sum −12, (−4)(−5) sum −9. The pair is −4 and −5.
x² − 9x + 20 = (x − 4)(x − 5)
Check: x² − 5x − 4x + 20 = x² − 9x + 20.
Answer: (x-4)*(x-5)
Factor: x² + 4x − 21.
Worked solutionTry it first, then open.
Two numbers multiply to c = −21 and add to b = +4. c < 0 ⇒ opposite signs; the number with the larger absolute value carries the sign of b (positive). Pairs of 21: 1·21, 3·7. We need a difference of 4 with the bigger one positive: +7 and −3 (product −21, sum +4).
x² + 4x − 21 = (x + 7)(x − 3)
Check: x² − 3x + 7x − 21 = x² + 4x − 21.
Answer: (x+7)*(x-3)
Factorable or prime? Decide for x² + 5x + 11 by listing the factor pairs out loud. (Light stretch — apply the stopping rule.)
Worked solutionTry it first, then open.
Need two numbers multiplying to c = 11 and adding to b = 5. c > 0 and b > 0 ⇒ both positive. 11 is prime, so the only positive pair is 1·11, which sums to 12 — not 5. No other integer pair multiplies to 11. The finite search is exhausted with no pair summing to 5.
Verdict: x² + 5x + 11 is prime (irreducible over the integers). 'Prime' here means the completed search came up empty — not a giveup. We do not reach for fractions or decimals; over the integers there is no factorization, and nothing to multiply back.
Answer: prime (irreducible over the integers)
Lesson 11.3: Special patterns
Factor: x² − 36.
Worked solutionTry it first, then open.
Two terms with a minus between two perfect squares ⇒ difference of squares, a² − b² = (a + b)(a − b). Here a = √(x²) = x and b = √36 = 6.
x² − 36 = (x + 6)(x − 6)
Check: x² − 6x + 6x − 36 = x² − 36 (the middle terms cancel).
Note: x² + 36 (a plus) would be a sum of squares — prime over the integers; difference of squares needs the minus.
Answer: (x+6)*(x-6)
Factor: x² + 14x + 49.
Worked solutionTry it first, then open.
Both ends are perfect squares: √(x²) = x and √49 = 7. Test the middle: 2·x·7 = 14x — matches the middle term, so it's a perfect-square trinomial. The middle sign (+) becomes the sign inside.
x² + 14x + 49 = (x + 7)²
Check: (x + 7)(x + 7) = x² + 7x + 7x + 49 = x² + 14x + 49.
Answer: (x+7)**2
Factor: 25x² − 4. Mind the square roots of the coefficients.
Worked solutionTry it first, then open.
Difference of squares with coefficients. 25x² = (5x)² so a = 5x; 4 = 2² so b = 2.
25x² − 4 = (5x + 2)(5x − 2)
Check: 25x² − 10x + 10x − 4 = 25x² − 4.
Watch: √(25x²) is 5x, not 25x — writing (25x + 2)(25x − 2) would multiply back to 625x² − 4, wrong.
Answer: (5*x+2)*(5*x-2)
Factor completely: 5x² − 45. (Light stretch — GCF first, then look for a pattern.)
Worked solutionTry it first, then open.
Always try the GCF first. Both coefficients are divisible by 5: 5x² − 45 = 5(x² − 9). Now the inside x² − 9 is a difference of squares (a = x, b = 3): x² − 9 = (x + 3)(x − 3).
5x² − 45 = 5(x + 3)(x − 3)
Check: 5(x + 3)(x − 3) = 5(x² − 9) = 5x² − 45.
The 5 must stay out front — (x + 3)(x − 3) alone would only give x² − 9. Notice the difference of squares wasn't visible until the GCF came out: that's why GCF goes first.
Answer: 5*(x+3)*(x-3)
Mixed review
Problems that mix skills from across the unit — good for spacing earlier work back in.
Factor out the GCF: 18x² + 24x.
Worked solutionTry it first, then open.
Numbers: gcd(18, 24) = 6. Variables: both terms have an x, lowest power x¹. GCF = 6x.
Divide: 18x² ÷ 6x = 3x, and 24x ÷ 6x = 4.
18x² + 24x = 6x(3x + 4)
Check: 6x·3x + 6x·4 = 18x² + 24x. (Stopping at 3x(6x + 8) or 2x(9x + 12) isn't the greatest — the inside still shares a factor.)
Answer: 6*x*(3*x+4)
Factor: x² − 3x − 18.
Worked solutionTry it first, then open.
Two numbers multiply to c = −18 and add to b = −3. c < 0 ⇒ opposite signs; the larger-absolute-value number carries b's sign (negative). Pairs of 18: 1·18, 2·9, 3·6. We need a difference of 3 with the bigger one negative: −6 and +3 (product −18, sum −3).
x² − 3x − 18 = (x − 6)(x + 3)
Check: x² + 3x − 6x − 18 = x² − 3x − 18.
Answer: (x-6)*(x+3)
Factor: x² − 64.
Worked solutionTry it first, then open.
Two terms, a minus between two perfect squares ⇒ difference of squares. a = √(x²) = x, b = √64 = 8.
x² − 64 = (x + 8)(x − 8)
Check: x² − 8x + 8x − 64 = x² − 64.
Answer: (x+8)*(x-8)
Factorable or prime? Decide for x² + x + 7 by listing the factor pairs.
Worked solutionTry it first, then open.
Need two numbers multiplying to c = 7 and adding to b = 1. c > 0 and b > 0 ⇒ both positive. 7 is prime, so the only positive pair is 1·7, summing to 8 — not 1. No other integer pair multiplies to 7. Search exhausted.
Verdict: x² + x + 7 is prime (irreducible over the integers). The completed pair list is the proof; there is no integer factorization to multiply back.
Answer: prime (irreducible over the integers)
Factor completely: 3x² − 30x + 75. (Stretch — pull the GCF, then check the leftover for a pattern.)
Worked solutionTry it first, then open.
GCF first: every coefficient (3, 30, 75) is divisible by 3, so 3x² − 30x + 75 = 3(x² − 10x + 25). Now look at the inside: both ends are perfect squares (√(x²) = x, √25 = 5) and the middle 2·x·5 = 10x matches — a perfect-square trinomial with a minus, so x² − 10x + 25 = (x − 5)².
3x² − 30x + 75 = 3(x − 5)²
Check: 3(x − 5)² = 3(x² − 10x + 25) = 3x² − 30x + 75.
Keep the 3 out front — it's part of the complete factorization.
Answer: 3*(x-5)**2