Factoring Polynomials worksheet with 12 algebraic expressions for practice.
Worksheet titled "Factoring Polynomials" with 12 polynomial expressions to factor, including quadratic and higher-degree polynomials, from Math Monks.
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Step-by-step solution for: Factoring Polynomials Worksheets with Answer Key
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Show Answer Key & Explanations
Step-by-step solution for: Factoring Polynomials Worksheets with Answer Key
Let’s solve each problem one by one. We’ll factor each polynomial completely.
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Problem 1: 3x² - 2x - 5
We need two numbers that multiply to (3)(-5) = -15 and add to -2.
Those numbers are -5 and +3.
Rewrite middle term:
3x² - 5x + 3x - 5
Group:
(3x² - 5x) + (3x - 5)
= x(3x - 5) + 1(3x - 5)
= (x + 1)(3x - 5)
✔ Final Answer for #1: (x + 1)(3x - 5)
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Problem 2: 10m² + 17m + 6
Multiply 10 × 6 = 60. Find two numbers that multiply to 60 and add to 17 → 12 and 5.
Rewrite:
10m² + 12m + 5m + 6
Group:
(10m² + 12m) + (5m + 6)
= 2m(5m + 6) + 1(5m + 6)
= (2m + 1)(5m + 6)
✔ Final Answer for #2: (2m + 1)(5m + 6)
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Problem 3: 2m² + 63m + 145
Multiply 2 × 145 = 290. Need two numbers that multiply to 290 and add to 63.
Try factors of 290:
1×290, 2×145, 5×58, 10×29 → 10 + 29 = 39 ✘
Wait — 58 + 5 = 63 ✔
So rewrite:
2m² + 58m + 5m + 145
Group:
(2m² + 58m) + (5m + 145)
= 2m(m + 29) + 5(m + 29)
= (2m + 5)(m + 29)
✔ Final Answer for #3: (2m + 5)(m + 29)
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Problem 4: 3x⁷ + 36x⁵ + 108x³
First, factor out GCF: all terms have 3x³.
= 3x³(x⁴ + 12x² + 36)
Now look at inside: x⁴ + 12x² + 36 → this is a perfect square trinomial!
Let u = x² → u² + 12u + 36 = (u + 6)²
So back: (x² + 6)²
Thus: 3x³(x² + 6)²
✔ Final Answer for #4: 3x³(x² + 6)²
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Problem 5: 12v² - 4v - 16
Factor out GCF first: 4
= 4(3v² - v - 4)
Now factor 3v² - v - 4
Multiply 3 × (-4) = -12. Need two numbers that multiply to -12 and add to -1 → -4 and +3
Rewrite:
3v² - 4v + 3v - 4
Group:
(3v² - 4v) + (3v - 4)
= v(3v - 4) + 1(3v - 4)
= (v + 1)(3v - 4)
Don’t forget the 4 we factored out!
✔ Final Answer for #5: 4(v + 1)(3v - 4)
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Problem 6: 3x² - 8x + 4
Multiply 3 × 4 = 12. Need two numbers that multiply to 12 and add to -8 → -6 and -2
Rewrite:
3x² - 6x - 2x + 4
Group:
(3x² - 6x) + (-2x + 4)
= 3x(x - 2) -2(x - 2)
= (3x - 2)(x - 2)
✔ Final Answer for #6: (3x - 2)(x - 2)
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Problem 7: 10u³ - 130u² + 400u
Factor out GCF: 10u
= 10u(u² - 13u + 40)
Now factor u² - 13u + 40 → find two numbers that multiply to 40 and add to -13 → -8 and -5
= 10u(u - 8)(u - 5)
✔ Final Answer for #7: 10u(u - 8)(u - 5)
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Problem 8: 121a² - 66ab + 9b²
This looks like a perfect square trinomial.
Check:
√(121a²) = 11a
√(9b²) = 3b
Middle term should be 2*(11a)*(3b) = 66ab → but here it’s -66ab → so sign is negative.
So: (11a - 3b)²
Check: (11a - 3b)² = 121a² - 66ab + 9b² ✔
✔ Final Answer for #8: (11a - 3b)²
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Problem 9: 4w² - 17w - 15
Multiply 4 × (-15) = -60. Need two numbers that multiply to -60 and add to -17 → -20 and +3
Rewrite:
4w² - 20w + 3w - 15
Group:
(4w² - 20w) + (3w - 15)
= 4w(w - 5) + 3(w - 5)
= (4w + 3)(w - 5)
✔ Final Answer for #9: (4w + 3)(w - 5)
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Problem 10: 2k² + 19k - 10
Multiply 2 × (-10) = -20. Need two numbers that multiply to -20 and add to 19 → 20 and -1
Rewrite:
2k² + 20k - k - 10
Group:
(2k² + 20k) + (-k - 10)
= 2k(k + 10) -1(k + 10)
= (2k - 1)(k + 10)
✔ Final Answer for #10: (2k - 1)(k + 10)
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Problem 11: 2m² + 63m + 145
Wait — this is identical to Problem #3! So same answer.
From #3: (2m + 5)(m + 29)
✔ Final Answer for #11: (2m + 5)(m + 29)
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Problem 12: 15a² + 45ab + 60ab²
Wait — let’s check the expression again:
It says “15a² + 45ab + 60ab²”
But note: last term has ab² — different variable power. Let’s write clearly:
Terms:
15a²
+ 45ab
+ 60ab²
All terms have ‘a’ and coefficient divisible by 15? Let’s see GCF.
Coefficients: 15, 45, 60 → GCF = 15
Variables: all have at least one ‘a’ → so factor out 15a
= 15a(a + 3b + 4b²)
Wait — let’s divide each term:
15a² ÷ 15a = a
45ab ÷ 15a = 3b
60ab² ÷ 15a = 4b²
So: 15a(a + 3b + 4b²)
Can we factor inside? a + 3b + 4b² → not really, since ‘a’ is separate from b terms. No common grouping.
So final factored form is 15a(4b² + 3b + a) — usually we write in order of powers of b or alphabetically.
Better to write as: 15a(a + 3b + 4b²) or rearrange: 15a(4b² + 3b + a)
Either is acceptable, but let’s write in standard polynomial order by degree of b:
→ 15a(4b² + 3b + a)
✔ Final Answer for #12: 15a(4b² + 3b + a)
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Final Answers:
1. (x + 1)(3x - 5)
2. (2m + 1)(5m + 6)
3. (2m + 5)(m + 29)
4. 3x³(x² + 6)²
5. 4(v + 1)(3v - 4)
6. (3x - 2)(x - 2)
7. 10u(u - 8)(u - 5)
8. (11a - 3b)²
9. (4w + 3)(w - 5)
10. (2k - 1)(k + 10)
11. (2m + 5)(m + 29)
12. 15a(4b² + 3b + a)
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Problem 1: 3x² - 2x - 5
We need two numbers that multiply to (3)(-5) = -15 and add to -2.
Those numbers are -5 and +3.
Rewrite middle term:
3x² - 5x + 3x - 5
Group:
(3x² - 5x) + (3x - 5)
= x(3x - 5) + 1(3x - 5)
= (x + 1)(3x - 5)
✔ Final Answer for #1: (x + 1)(3x - 5)
---
Problem 2: 10m² + 17m + 6
Multiply 10 × 6 = 60. Find two numbers that multiply to 60 and add to 17 → 12 and 5.
Rewrite:
10m² + 12m + 5m + 6
Group:
(10m² + 12m) + (5m + 6)
= 2m(5m + 6) + 1(5m + 6)
= (2m + 1)(5m + 6)
✔ Final Answer for #2: (2m + 1)(5m + 6)
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Problem 3: 2m² + 63m + 145
Multiply 2 × 145 = 290. Need two numbers that multiply to 290 and add to 63.
Try factors of 290:
1×290, 2×145, 5×58, 10×29 → 10 + 29 = 39 ✘
Wait — 58 + 5 = 63 ✔
So rewrite:
2m² + 58m + 5m + 145
Group:
(2m² + 58m) + (5m + 145)
= 2m(m + 29) + 5(m + 29)
= (2m + 5)(m + 29)
✔ Final Answer for #3: (2m + 5)(m + 29)
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Problem 4: 3x⁷ + 36x⁵ + 108x³
First, factor out GCF: all terms have 3x³.
= 3x³(x⁴ + 12x² + 36)
Now look at inside: x⁴ + 12x² + 36 → this is a perfect square trinomial!
Let u = x² → u² + 12u + 36 = (u + 6)²
So back: (x² + 6)²
Thus: 3x³(x² + 6)²
✔ Final Answer for #4: 3x³(x² + 6)²
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Problem 5: 12v² - 4v - 16
Factor out GCF first: 4
= 4(3v² - v - 4)
Now factor 3v² - v - 4
Multiply 3 × (-4) = -12. Need two numbers that multiply to -12 and add to -1 → -4 and +3
Rewrite:
3v² - 4v + 3v - 4
Group:
(3v² - 4v) + (3v - 4)
= v(3v - 4) + 1(3v - 4)
= (v + 1)(3v - 4)
Don’t forget the 4 we factored out!
✔ Final Answer for #5: 4(v + 1)(3v - 4)
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Problem 6: 3x² - 8x + 4
Multiply 3 × 4 = 12. Need two numbers that multiply to 12 and add to -8 → -6 and -2
Rewrite:
3x² - 6x - 2x + 4
Group:
(3x² - 6x) + (-2x + 4)
= 3x(x - 2) -2(x - 2)
= (3x - 2)(x - 2)
✔ Final Answer for #6: (3x - 2)(x - 2)
---
Problem 7: 10u³ - 130u² + 400u
Factor out GCF: 10u
= 10u(u² - 13u + 40)
Now factor u² - 13u + 40 → find two numbers that multiply to 40 and add to -13 → -8 and -5
= 10u(u - 8)(u - 5)
✔ Final Answer for #7: 10u(u - 8)(u - 5)
---
Problem 8: 121a² - 66ab + 9b²
This looks like a perfect square trinomial.
Check:
√(121a²) = 11a
√(9b²) = 3b
Middle term should be 2*(11a)*(3b) = 66ab → but here it’s -66ab → so sign is negative.
So: (11a - 3b)²
Check: (11a - 3b)² = 121a² - 66ab + 9b² ✔
✔ Final Answer for #8: (11a - 3b)²
---
Problem 9: 4w² - 17w - 15
Multiply 4 × (-15) = -60. Need two numbers that multiply to -60 and add to -17 → -20 and +3
Rewrite:
4w² - 20w + 3w - 15
Group:
(4w² - 20w) + (3w - 15)
= 4w(w - 5) + 3(w - 5)
= (4w + 3)(w - 5)
✔ Final Answer for #9: (4w + 3)(w - 5)
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Problem 10: 2k² + 19k - 10
Multiply 2 × (-10) = -20. Need two numbers that multiply to -20 and add to 19 → 20 and -1
Rewrite:
2k² + 20k - k - 10
Group:
(2k² + 20k) + (-k - 10)
= 2k(k + 10) -1(k + 10)
= (2k - 1)(k + 10)
✔ Final Answer for #10: (2k - 1)(k + 10)
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Problem 11: 2m² + 63m + 145
Wait — this is identical to Problem #3! So same answer.
From #3: (2m + 5)(m + 29)
✔ Final Answer for #11: (2m + 5)(m + 29)
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Problem 12: 15a² + 45ab + 60ab²
Wait — let’s check the expression again:
It says “15a² + 45ab + 60ab²”
But note: last term has ab² — different variable power. Let’s write clearly:
Terms:
15a²
+ 45ab
+ 60ab²
All terms have ‘a’ and coefficient divisible by 15? Let’s see GCF.
Coefficients: 15, 45, 60 → GCF = 15
Variables: all have at least one ‘a’ → so factor out 15a
= 15a(a + 3b + 4b²)
Wait — let’s divide each term:
15a² ÷ 15a = a
45ab ÷ 15a = 3b
60ab² ÷ 15a = 4b²
So: 15a(a + 3b + 4b²)
Can we factor inside? a + 3b + 4b² → not really, since ‘a’ is separate from b terms. No common grouping.
So final factored form is 15a(4b² + 3b + a) — usually we write in order of powers of b or alphabetically.
Better to write as: 15a(a + 3b + 4b²) or rearrange: 15a(4b² + 3b + a)
Either is acceptable, but let’s write in standard polynomial order by degree of b:
→ 15a(4b² + 3b + a)
✔ Final Answer for #12: 15a(4b² + 3b + a)
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Final Answers:
1. (x + 1)(3x - 5)
2. (2m + 1)(5m + 6)
3. (2m + 5)(m + 29)
4. 3x³(x² + 6)²
5. 4(v + 1)(3v - 4)
6. (3x - 2)(x - 2)
7. 10u(u - 8)(u - 5)
8. (11a - 3b)²
9. (4w + 3)(w - 5)
10. (2k - 1)(k + 10)
11. (2m + 5)(m + 29)
12. 15a(4b² + 3b + a)
Parent Tip: Review the logic above to help your child master the concept of free factoring worksheet.