Factoring Polynomials worksheet with 12 algebraic expressions to solve.

Let’s solve each problem step by step. We’re factoring polynomials — that means writing them as a product of simpler expressions.

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1. 3x² - 2x - 5

We look for two numbers that multiply to (3)(-5) = -15 and add to -2.
Those numbers are -5 and +3.

Split the middle term:
3x² - 5x + 3x - 5
Group: (3x² - 5x) + (3x - 5)
Factor each group: x(3x - 5) + 1(3x - 5)
→ (x + 1)(3x - 5)

✔ Check: (x+1)(3x-5) = 3x² -5x +3x -5 = 3x² -2x -5 ✔️

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2. 10m² + 17m + 6

Multiply 10 × 6 = 60. Find two numbers that multiply to 60 and add to 17 → 12 and 5.

Split: 10m² + 12m + 5m + 6
Group: (10m² + 12m) + (5m + 6)
Factor: 2m(5m + 6) + 1(5m + 6)
→ (2m + 1)(5m + 6)

✔ Check: (2m+1)(5m+6) = 10m² +12m +5m +6 = 10m² +17m +6 ✔️

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3. 2m² + 63m + 145

Multiply 2 × 145 = 290. Need two numbers that multiply to 290 and add to 63.

Factors of 290: 1×290, 2×145, 5×58, 10×29 → 10 and 29? 10+29=39 no. 5+58=63 YES!

So split: 2m² + 5m + 58m + 145
Group: (2m² + 5m) + (58m + 145)
Factor: m(2m + 5) + 29(2m + 5)
→ (m + 29)(2m + 5)

✔ Check: (m+29)(2m+5) = 2m² +5m +58m +145 = 2m² +63m +145 ✔️

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4. 3x⁷ + 36x⁵ + 108x³

First, factor out GCF: all terms have 3x³.

→ 3x³(x⁴ + 12x² + 36)

Now factor inside: x⁴ + 12x² + 36 is a quadratic in form. Let u = x² → u² + 12u + 36 = (u+6)²

So → 3x³(x² + 6)²

✔ Check: 3x³(x²+6)² = 3x³(x⁴ +12x² +36) = 3x⁷ +36x⁵ +108x³ ✔️

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5. 12v² - 4v - 16

GCF first: 4 divides all → 4(3v² - v - 4)

Now factor 3v² - v - 4. Multiply 3×(-4)= -12. Numbers that multiply to -12 and add to -1 → -4 and +3.

Split: 3v² -4v +3v -4
Group: (3v² -4v) + (3v -4) → v(3v -4) +1(3v -4) → (v+1)(3v-4)

So overall: 4(v + 1)(3v - 4)

✔ Check: 4(v+1)(3v-4) = 4[3v² -4v +3v -4] = 4[3v² -v -4] = 12v² -4v -16 ✔️

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6. 3x² - 8x + 4

Multiply 3×4=12. Need two numbers that multiply to 12 and add to -8 → -6 and -2.

Split: 3x² -6x -2x +4
Group: (3x² -6x) + (-2x +4) → 3x(x - 2) -2(x - 2) → (3x - 2)(x - 2)

✔ Check: (3x-2)(x-2) = 3x² -6x -2x +4 = 3x² -8x +4 ✔️

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7. 10u³ - 130u² + 400u

GCF: 10u → 10u(u² - 13u + 40)

Factor u² -13u +40: need two numbers that multiply to 40 and add to -13 → -8 and -5.

→ (u - 8)(u - 5)

So overall: 10u(u - 8)(u - 5)

✔ Check: 10u(u-8)(u-5) = 10u[u² -13u +40] = 10u³ -130u² +400u ✔️

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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 (11a - 3b)²

→ (11a - 3b)²

✔ Check: (11a - 3b)² = 121a² - 66ab + 9b² ✔️

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9. 4w² - 17w - 15

Multiply 4×(-15) = -60. Need two numbers that multiply to -60 and add to -17 → -20 and +3.

Split: 4w² -20w +3w -15
Group: (4w² -20w) + (3w -15) → 4w(w - 5) +3(w - 5) → (4w + 3)(w - 5)

✔ Check: (4w+3)(w-5) = 4w² -20w +3w -15 = 4w² -17w -15 ✔️

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10. 2k² + 19k - 10

Multiply 2×(-10) = -20. Need two numbers that multiply to -20 and add to 19 → 20 and -1.

Split: 2k² +20k -k -10
Group: (2k² +20k) + (-k -10) → 2k(k + 10) -1(k + 10) → (2k - 1)(k + 10)

✔ Check: (2k-1)(k+10) = 2k² +20k -k -10 = 2k² +19k -10 ✔️

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11. 2m² + 63m + 145

Wait — this is the same as #3! So answer is also (m + 29)(2m + 5)

But let’s double-check just in case.

Same steps: 2×145=290; factors 5 and 58 → 5+58=63 → yes.

→ (m + 29)(2m + 5)

✔ Same as #3.

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12. 15a² + 45ab + 60ab²

Wait — this seems odd. Terms are: 15a², 45ab, 60ab².

Note: The last term has b², others don’t. But we can still factor.

First, GCF: All coefficients divisible by 15? 15, 45, 60 → yes. Variables: all have at least ‘a’? First term: a², second: ab, third: ab² → common variable is ‘a’.

So GCF = 15a

Factor out: 15a(a + 3b + 4b²)

Wait — inside: a + 3b + 4b² → rearrange: 4b² + 3b + a → not factorable further since ‘a’ is alone.

Actually, maybe typo? Original says “60ab²” — perhaps meant “60b²”? But as written, we proceed.

Alternatively, maybe it’s 15a² + 45ab + 60b²? That would make more sense for factoring.

But user wrote: 15a² + 45ab + 60ab²

Let me check again: 15a² + 45ab + 60ab²

Factor 15a: → 15a(a + 3b + 4b²)

Inside: a + 3b + 4b² — cannot factor further because ‘a’ is linear and doesn’t match with b terms.

But wait — maybe it’s supposed to be 15a² + 45ab + 60b²? That would be standard.

Looking back at image description — user said “12 15a² + 45ab + 60ab²”

Hmm. Perhaps it's a misprint? In many textbooks, it might be 60b².

But if we take it literally: 15a² + 45ab + 60ab² = 15a(a + 3b + 4b²)

But that feels incomplete. Alternatively, maybe group differently?

Try grouping: 15a² + 45ab + 60ab² = 15a² + 45ab + 60ab²

Factor 15a from first two: 15a(a + 3b) + 60ab² — not helpful.

Alternatively, factor 15a from all: as above.

Perhaps the intended expression was 15a² + 45ab + 60b²? Let’s assume that for a moment — because otherwise it doesn't factor nicely.

If it were 15a² + 45ab + 60b²:

GCF = 15 → 15(a² + 3ab + 4b²) — still doesn’t factor over integers.

Wait — 15a² + 45ab + 60b² = 15(a² + 3ab + 4b²) — discriminant for quadratic in a: (3b)^2 - 4*1*4b² = 9b² - 16b² = -7b² <0 → no real roots → doesn’t factor.

But original has “60ab²” — which includes an extra ‘a’. So perhaps it’s correct as is.

Another thought: Maybe it’s 15a² + 45ab + 60b²? Or 15a² + 45ab + 60a b^2 — same thing.

Wait — let’s try factoring as is:

15a² + 45ab + 60ab² = 15a(a + 3b + 4b²)

And 4b² + 3b + a — no common factors, doesn’t factor further.

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

I think we have to go with that.

But let me check if there’s another way.

Suppose we treat it as polynomial in a: 15a² + (45b + 60b²)a + 0 — no constant term? Wait, no constant term, but we already factored 15a.

Actually, after factoring 15a, we have left: a + 3b + 4b² — which is fine.

So final answer: 15a(4b² + 3b + a) or 15a(a + 3b + 4b²)

Usually we write in order of powers: 15a(4b² + 3b + a)

But typically we arrange variables alphabetically or by degree. Since ‘a’ is degree 1, ‘b’ up to 2, perhaps leave as is.

To match standard form, write inside as 4b² + 3b + a.

But I recall — in some cases, if it’s 60ab², maybe it’s a typo and should be 60b². However, since the problem says “60ab²”, we’ll keep it.

Wait — let me calculate numerically to verify.

Suppose a=1, b=1: original = 15(1) + 45(1)(1) + 60(1)(1) = 15+45+60=120

Our factored: 15*1*(1 + 3*1 + 4*1²) = 15*(1+3+4)=15*8=120 ✔️

Another point: a=2, b=1: original = 15*4 + 45*2*1 + 60*2*1 = 60 + 90 + 120 = 270

Factored: 15*2*(2 + 3*1 + 4*1) = 30*(2+3+4)=30*9=270 ✔️

So it’s correct.

Thus, 15a(a + 3b + 4b²)

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

In standard form, often we write higher degree first: 15a(4b² + 3b + a)

But since ‘a’ is a different variable, it’s acceptable.

Some might prefer to write the polynomial in descending powers of one variable, but here it’s mixed.

I think 15a(a + 3b + 4b²) is fine.

But let’s see if we can factor the inside — 4b² + 3b + a — no, because ‘a’ is not related to b.

So final answer for #12: 15a(a + 3b + 4b²)

However, looking back at the original list, #12 is "15a² + 45ab + 60ab²" — which is unusual, but mathematically correct as factored.

Perhaps the intention was 15a² + 45ab + 60b², but even then it doesn’t factor nicely.

Another possibility: maybe it’s 15a² + 45ab + 60b², and we factor 15: 15(a² + 3ab + 4b²) — still prime.

Or if it were 15a² + 45ab + 30b², etc.

But given what’s written, we stick with 15a(a + 3b + 4b²)

To make it neat, write as 15a(4b² + 3b + a)

But in most textbooks, they might expect to factor completely, and this is complete.

So I'll go with that.

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Now, compiling all answers:

Final Answer:

1. (x + 1)(3x - 5)
2. (2m + 1)(5m + 6)
3. (m + 29)(2m + 5)
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. (m + 29)(2m + 5)
12. 15a(a + 3b + 4b²)

Note: For #12, if the problem intended "60b²" instead of "60ab²", the answer would be different, but based on the given expression, this is correct.