Factoring Polynomials Worksheets with Answer Key - Free Printable

Let’s solve each problem one by one. We’re factoring polynomials — that means rewriting them as a product of simpler expressions (like breaking apart numbers into factors, but with variables).

We’ll go step by step for each.

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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: 58 and 5 → 58×5=290, 58+5=63 ✔

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 the GCF: all terms have 3 and x³ → 3x³

So:
3x³(x⁴ + 12x² + 36)

Now look at inside: x⁴ + 12x² + 36 → this is a perfect square trinomial!

It’s (x²)² + 2·x²·6 + 6² = (x² + 6)²

So overall:
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: all divisible by 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 pulled 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: all divisible by 10u → 10u(u² - 13u + 40)

Now factor u² - 13u + 40 → find two numbers that multiply to 40 and add to -13 → -8 and -5.

So: (u - 8)(u - 5)

Overall: 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 → yes, and it’s negative → so (11a - 3b)²

✔ 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 the same as Problem #3! So answer is the same.

✔ Final Answer for #11: (2m + 5)(m + 29)

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Problem 12: 15a² + 45ab + 60ab²

Wait — let’s check the expression: 15a² + 45ab + 60ab²

Actually, looking closely — maybe there’s a typo? Because 60ab² has b², while others don’t. But let’s assume it’s written correctly.

But notice: all terms have “a” and also divisible by 15? Let’s see:

GCF of coefficients: 15, 45, 60 → GCF = 15
Each term has at least one ‘a’ → so factor out 15a

So:
15a(a + 3b + 4b²)

Wait — that gives: 15a·a = 15a², 15a·3b = 45ab, 15a·4b² = 60ab² → correct.

But can we factor further? Inside: a + 3b + 4b² — not really factorable since it’s mixed variables and no common structure.

Alternatively — maybe the original was meant to be 15a² + 45ab + 60b²? That would make more sense for factoring.

But based on what’s written: 15a² + 45ab + 60ab²

We can write it as: 15a(a + 3b + 4b²)

Or rearrange inside: 15a(4b² + 3b + a) — still not factorable nicely.

But wait — perhaps the last term is 60b²? Let me double-check the image description... In the user input, it says: "12 15a² + 45ab + 60ab²"

Hmm. If it’s truly 60ab², then factoring out 15a is best.

But let’s consider: maybe it’s a typo and should be 60b²? Because otherwise, it’s unusual.

In many textbooks, problems like this are symmetric or follow patterns.

Assuming it’s 15a² + 45ab + 60b² — then:

Factor out 15: 15(a² + 3ab + 4b²) — but a² + 3ab + 4b² doesn’t factor nicely.

Wait — 15a² + 45ab + 60b² → divide by 15: a² + 3ab + 4b² — discriminant = 9b² - 16b² = -7b² → not real factors.

Alternatively, if it’s 15a² + 45ab + 30b²? Not matching.

Wait — another thought: maybe it’s 15a² + 45ab + 60ab² — but that’s asymmetric.

Perhaps group differently?

15a² + 45ab + 60ab² = 15a² + 45ab + 60ab²

Factor 15a from first two? No.

Actually, let’s factor 15a from all? As before: 15a(a + 3b + 4b²)

That’s acceptable.

But let me check if 4b² + 3b + a can be factored — no, because it’s in terms of b and a mixed.

So safest answer: factor out GCF 15a.

✔ Final Answer for #12: 15a(a + 3b + 4b²)

But wait — perhaps the problem meant 15a² + 45ab + 60b²? Then we’d factor 15: 15(a² + 3ab + 4b²) — which doesn’t factor.

Alternatively, if it’s 15a² + 45ab + 30b²? Still not nice.

Another possibility: maybe it’s 15a² + 45ab + 60b² — and we factor 15, then try to factor quadratic in a: a² + 3b a + 4b² — discriminant 9b² - 16b² = -7b² — no real roots.

So I think the intended expression might have been different, but based on what’s given, we go with factoring out 15a.

However, looking back at the original list — problem 12 is: “15a² + 45ab + 60ab²”

Notice: 60ab² — that’s 60 * a * b²

So terms are: 15a², 45ab, 60ab²

All have ‘a’, and coefficients divisible by 15.

So: 15a(a + 3b + 4b²)

Yes.

We can write it as 15a(4b² + 3b + a) — same thing.

I think that’s the best we can do.

✔ Final Answer for #12: 15a(a + 3b + 4b²)

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Now, compiling all final answers:

Final Answer:
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(a + 3b + 4b²)