➕ Algebra 1 · Factoring
Factoring Trinomials practice questions
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Factor x2 + 12x + 35.
Answer: (x + 5)(x + 7). Look for two integers with product 35 and sum 12. Both are positive, because c and b are positive. The pairs with product 35 are 1 · 35 (sum 36) and 5 · 7 (sum 12), so the answer is (x + 5)(x + 7). Check: x2 + 7x + 5x + 35 = x2 + 12x + 35. The choice (x − 5)(x − 7) has the right product but sum −12. The choice (x + 1)(x + 35) has product 35 but sum 36. The choice (x + 6)(x + 6) has sum 12 but product 36.
Factor x2 − 8x + 12.
Answer: (x − 2)(x − 6). Since c = 12 is positive and b = −8 is negative, both numbers are negative. The pairs with product 12 are (−1)(−12) (sum −13), (−2)(−6) (sum −8) and (−3)(−4) (sum −7). So the answer is (x − 2)(x − 6). Check: x2 − 6x − 2x + 12 = x2 − 8x + 12. The choice (x + 2)(x + 6) would give +8x. The choice (x − 3)(x − 4) has product 12 but sum −7. The choice (x − 2)(x + 6) has sum 4 and product −12.
Factor x2 + 3x − 28.
Answer: (x + 7)(x − 4). Since c = −28 is negative, the numbers have opposite signs, and since b = 3 is positive, the larger one is positive. The pairs with product −28 are (−1)(28) (sum 27), (−2)(14) (sum 12) and (−4)(7) (sum 3). So the answer is (x + 7)(x − 4). Check: x2 − 4x + 7x − 28 = x2 + 3x − 28. The choice (x − 7)(x + 4) has sum −3. The choice (x + 7)(x + 4) has product +28. The choice (x + 14)(x − 2) has product −28 but sum 12.
Factor 4x2 − 28x + 48 completely.
Answer: 4(x − 3)(x − 4). Take out the GCF, 4, first: 4(x2 − 7x + 12). Inside, c = 12 is positive and b = −7 is negative, so both numbers are negative: (−3)(−4) = 12 and −3 + (−4) = −7. So the answer is 4(x − 3)(x − 4). Check: (x − 3)(x − 4) = x2 − 7x + 12, and times 4 gives 4x2 − 28x + 48. The choice (x − 3)(x − 4) drops the GCF. The choice (4x − 12)(x − 4) expands correctly but is not complete, because 4x − 12 still has the common factor 4. The choice 4(x + 3)(x + 4) has the wrong signs.
Factor −x2 − 2x + 24 completely by taking out −1 first.
Answer: −(x + 6)(x − 4). Factor out −1, which flips every sign inside: −(x2 + 2x − 24). Inside, c = −24 is negative, so the numbers have opposite signs, and b = 2 is positive, so the larger one is positive. The pairs with product −24 are (−1)(24) (sum 23), (−2)(12) (sum 10), (−3)(8) (sum 5) and (−4)(6) (sum 2). So x2 + 2x − 24 = (x + 6)(x − 4), and the answer is −(x + 6)(x − 4). Check: (x + 6)(x − 4) = x2 + 2x − 24, and times −1 gives −x2 − 2x + 24. The choice (x + 6)(x − 4) forgets the −1. The choice −(x − 6)(x + 4) swaps the signs of the pair and expands to −x2 + 2x + 24. The choice −(x + 6)(x + 4) expands to −x2 − 10x − 24.
Factor 3x2 + 14x + 8.
Answer: (3x + 2)(x + 4). There is no GCF. Here ac = 3 · 8 = 24 and b = 14, so look for two positive integers with product 24 and sum 14. The pairs are 1 · 24 (sum 25), 2 · 12 (sum 14), 3 · 8 (sum 11) and 4 · 6 (sum 10). The pair is 2 and 12. Split the middle term: 3x2 + 2x + 12x + 8 = x(3x + 2) + 4(3x + 2) = (3x + 2)(x + 4). Check: 3x2 + 12x + 2x + 8 = 3x2 + 14x + 8. The choice (3x + 4)(x + 2) expands to 3x2 + 10x + 8, the choice (3x + 8)(x + 1) to 3x2 + 11x + 8, and the choice (3x − 2)(x − 4) to 3x2 − 14x + 8.
Factor 8x2 + 2x − 15.
Answer: (2x + 3)(4x − 5). There is no GCF. Here ac = 8 · (−15) = −120 and b = 2, so look for two integers with product −120 and sum 2. They have opposite signs and the larger one is positive. Trying the pairs, (−10)(12) has product −120 and sum 2, so the pair is −10 and 12. Split the middle term: 8x2 + 12x − 10x − 15 = 4x(2x + 3) − 5(2x + 3) = (2x + 3)(4x − 5). Check: 8x2 − 10x + 12x − 15 = 8x2 + 2x − 15. The choice (2x − 3)(4x + 5) expands to 8x2 − 2x − 15. The choice (2x + 3)(4x + 5) expands to 8x2 + 22x + 15. The choice (2x + 5)(4x − 3) expands to 8x2 + 14x − 15.
Factor 6x2 + 21x + 15 completely.
Answer: 3(2x + 5)(x + 1). Take out the GCF, 3, first: 3(2x2 + 7x + 5). Inside, ac = 2 · 5 = 10 and b = 7, so look for two positive integers with product 10 and sum 7. The pair is 2 and 5. Split the middle term: 2x2 + 2x + 5x + 5 = 2x(x + 1) + 5(x + 1) = (x + 1)(2x + 5). So the answer is 3(2x + 5)(x + 1). Check: (2x + 5)(x + 1) = 2x2 + 7x + 5, and times 3 gives 6x2 + 21x + 15. The choice (2x + 5)(x + 1) drops the GCF. The choice (6x + 15)(x + 1) expands correctly but is not complete, because 6x + 15 still has the common factor 3. The choice 3(2x + 1)(x + 5) expands to 6x2 + 33x + 15.
Which of these trinomials cannot be factored using integers (is prime)?
Answer: x2 + 4x + 9. Test each one. x2 + 9x + 8 = (x + 1)(x + 8), because 1 · 8 = 8 and 1 + 8 = 9. x2 − 2x − 8 = (x − 4)(x + 2), because (−4)(2) = −8 and −4 + 2 = −2. 2x2 + 7x + 3 = (2x + 1)(x + 3): here ac = 6, and 1 and 6 add to 7. For x2 + 4x + 9, the positive pairs with product 9 are 1 · 9 (sum 10) and 3 · 3 (sum 6), and negative pairs have negative sums. No pair adds to 4, so it is prime.
Factor 25x2 − 30x + 9.
Answer: (5x − 3)2. The first term is (5x)2 and the last term is 32. The middle term is −2 · 5x · 3 = −30x, so this is a perfect square trinomial with a minus sign: (5x − 3)2. Check: (5x − 3)(5x − 3) = 25x2 − 15x − 15x + 9 = 25x2 − 30x + 9. The choice (5x + 3)2 has a middle term of +30x. The choice (5x − 3)(5x + 3) expands to 25x2 − 9, with no middle term. The choice (5x − 9)(5x − 1) expands to 25x2 − 50x + 9.