Factoring Polynomials Worksheet 2 displaying algebraic expressions for practice in factoring using Difference and Sum of Squares formulas.

Let’s solve each problem step by step. Since this is a factoring worksheet, we’ll factor each polynomial completely using methods like GCF (greatest common factor), trinomial factoring, difference of squares, sum/difference of cubes, etc.

We’ll go one by one — but since there are 30 problems, I’ll group them logically and show the key steps for each type. But to keep it clear and accurate, I’ll solve all 30 with careful checking.

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Problem 1: 3p² – 2p – 5

Look for two numbers that multiply to 3 × (-5) = -15 and add to -2 → -5 and +3

Split middle term:
3p² + 3p – 5p – 5
= 3p(p + 1) –5(p + 1)
= (3p – 5)(p + 1)

✔ Check: (3p–5)(p+1) = 3p² +3p –5p –5 = 3p² –2p –5 ✔️

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Problem 2: 3x² – 6x – 45

First, factor out GCF: 3(x² – 2x – 15)

Now factor x² – 2x – 15 → numbers: -5 and +3 → (x – 5)(x + 3)

So: 3(x – 5)(x + 3)

✔ Check: 3(x–5)(x+3) = 3(x² –2x –15) = 3x² –6x –45 ✔️

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Problem 3: 4w² – 15w – 25

Multiply 4 × (-25) = -100; find two numbers that multiply to -100 and add to -15 → -20 and +5

Split: 4w² –20w +5w –25
= 4w(w – 5) +5(w – 5)
= (4w + 5)(w – 5)

✔ Check: (4w+5)(w–5) = 4w² –20w +5w –25 = 4w² –15w –25 ✔️

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Problem 4: 4y² – 20y + 25

This looks like a perfect square trinomial.

(2y)² = 4y², (5)² = 25, 2×2y×5 = 20y → yes!

So: (2y – 5)²

✔ Check: (2y–5)² = 4y² –20y +25 ✔️

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Problem 5: 10j³ – 130j² + 400j

Factor out GCF: 10j(j² – 13j + 40)

Factor j² –13j +40 → numbers: -8 and -5 → (j – 8)(j – 5)

So: 10j(j – 8)(j – 5)

✔ Check: 10j(j–8)(j–5) = 10j(j² –13j +40) = 10j³ –130j² +400j ✔️

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Problem 6: –49 + 35b – 4b²

Rewrite in standard form: –4b² + 35b – 49

Factor out -1: -(4b² – 35b + 49)

Now factor 4b² –35b +49 → multiply 4×49=196; find two numbers that multiply to 196 and add to -35 → -28 and -7

Split: 4b² –28b –7b +49
= 4b(b – 7) –7(b – 7)
= (4b – 7)(b – 7)

So overall: -(4b – 7)(b – 7) OR (7 – 4b)(b – 7) — both acceptable, but usually write leading positive if possible.

Actually, better to write as: (7 – 4b)(b – 7) or leave as -(4b–7)(b–7). Let’s check:

-(4b–7)(b–7) = -[4b² –28b –7b +49] = -[4b² –35b +49] = -4b² +35b –49 ✔️

But original was –49 +35b –4b² → same thing.

Alternatively, rearrange factors: (7 – 4b)(b – 7) also works.

I’ll go with: (7 – 4b)(b – 7) — avoids negative sign outside.

Wait — let me verify:

(7 – 4b)(b – 7) = 7b – 49 –4b² +28b = -4b² +35b –49 ✔️

Yes. So final answer: (7 – 4b)(b – 7)

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Problem 7: 2n² + 5mn + 2m²

Treat as quadratic in n: 2n² + 5m n + 2m²

Multiply 2 × 2m² = 4m²; need two terms that multiply to 4m² and add to 5m → 4m and m

Split: 2n² + 4m n + m n + 2m²
= 2n(n + 2m) + m(n + 2m)
= (2n + m)(n + 2m)

✔ Check: (2n+m)(n+2m) = 2n² +4mn + mn + 2m² = 2n² +5mn +2m² ✔️

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Problem 8: 4y² – 17y + 4

Multiply 4×4=16; find two numbers that multiply to 16 and add to -17 → -16 and -1

Split: 4y² –16y – y + 4
= 4y(y – 4) –1(y – 4)
= (4y – 1)(y – 4)

✔ Check: (4y–1)(y–4) = 4y² –16y –y +4 = 4y² –17y +4 ✔️

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Problem 9: –5x² + 18x – 9

Factor out -1: -(5x² – 18x + 9)

Factor 5x² –18x +9 → multiply 5×9=45; numbers: -15 and -3

Split: 5x² –15x –3x +9
= 5x(x – 3) –3(x – 3)
= (5x – 3)(x – 3)

So overall: -(5x – 3)(x – 3) OR (3 – 5x)(x – 3)

Check: (3–5x)(x–3) = 3x –9 –5x² +15x = -5x² +18x –9 ✔️

So: (3 – 5x)(x – 3)

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Problem 10: 12u² – 4u – 16

GCF: 4 → 4(3u² – u – 4)

Factor 3u² – u – 4 → multiply 3×(-4)= -12; numbers: -4 and +3

Split: 3u² –4u +3u –4
= u(3u – 4) +1(3u – 4)
= (u + 1)(3u – 4)

So: 4(u + 1)(3u – 4)

✔ Check: 4(u+1)(3u–4) = 4[3u² –4u +3u –4] = 4[3u² –u –4] = 12u² –4u –16 ✔️

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Problem 11: z⁴ + 6z² – 7z²? Wait — typo?

Wait, problem says: z⁴ + 6z² – 7z²? That would be z⁴ – z².

But looking at image: probably z⁴ + 6z² – 7? Because otherwise it's too simple.

Wait — user wrote: “11 z⁴ + 6z² – 7z²” — that must be a typo in transcription. Looking back at original image description — actually, in the image it’s likely z⁴ + 6z² – 7, because 6z² –7z² doesn’t make sense.

Assuming it’s z⁴ + 6z² – 7

Let u = z² → u² + 6u – 7 → factors: (u + 7)(u – 1)

So: (z² + 7)(z² – 1) → then z² –1 is difference of squares → (z – 1)(z + 1)

Final: (z² + 7)(z – 1)(z + 1)

✔ Check: (z²+7)(z²–1) = z⁴ –z² +7z² –7 = z⁴ +6z² –7 ✔️

If it really was z⁴ +6z² –7z² = z⁴ – z² = z²(z² –1) = z²(z–1)(z+1) — but that seems unlikely given context.

Given the pattern, I think it’s z⁴ + 6z² – 7

So I’ll go with: (z² + 7)(z – 1)(z + 1)

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Problem 12: 9k² + 66k + 21

GCF: 3 → 3(3k² + 22k + 7)

Factor 3k² +22k +7 → multiply 3×7=21; numbers: 21 and 1

Split: 3k² +21k + k +7
= 3k(k + 7) +1(k + 7)
= (3k + 1)(k + 7)

So: 3(3k + 1)(k + 7)

✔ Check: 3(3k+1)(k+7) = 3[3k² +21k +k +7] = 3[3k² +22k +7] = 9k² +66k +21 ✔️

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Problem 13: 6x² – 7xy – 3y²

Treat as quadratic in x: 6x² –7y x –3y²

Multiply 6×(-3y²)= -18y²; find two terms: -9y and +2y (since -9y * 2y = -18y², -9y + 2y = -7y)

Split: 6x² –9yx +2yx –3y²
= 3x(2x – 3y) + y(2x – 3y)
= (3x + y)(2x – 3y)

✔ Check: (3x+y)(2x–3y) = 6x² –9xy +2xy –3y² = 6x² –7xy –3y² ✔️

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Problem 14: 6b² – 15b – 9

GCF: 3 → 3(2b² – 5b – 3)

Factor 2b² –5b –3 → multiply 2×(-3)= -6; numbers: -6 and +1

Split: 2b² –6b + b –3
= 2b(b – 3) +1(b – 3)
= (2b + 1)(b – 3)

So: 3(2b + 1)(b – 3)

✔ Check: 3(2b+1)(b–3) = 3[2b² –6b +b –3] = 3[2b² –5b –3] = 6b² –15b –9 ✔️

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Problem 15: 4w² – 17w – 15

Multiply 4×(-15)= -60; numbers: -20 and +3

Split: 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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Problem 16: 4y² – 17y + 15

Multiply 4×15=60; numbers: -12 and -5

Split: 4y² –12y –5y +15
= 4y(y – 3) –5(y – 3)
= (4y – 5)(y – 3)

✔ Check: (4y–5)(y–3) = 4y² –12y –5y +15 = 4y² –17y +15 ✔️

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Problem 17: 15m³ + 9m² – 6m

GCF: 3m → 3m(5m² + 3m – 2)

Factor 5m² +3m –2 → multiply 5×(-2)= -10; numbers: +5 and -2

Split: 5m² +5m –2m –2
= 5m(m + 1) –2(m + 1)
= (5m – 2)(m + 1)

So: 3m(5m – 2)(m + 1)

✔ Check: 3m(5m–2)(m+1) = 3m[5m² +5m –2m –2] = 3m[5m² +3m –2] = 15m³ +9m² –6m ✔️

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Problem 18: z⁴ + 2z² – 1

Let u = z² → u² + 2u – 1 → does not factor nicely over integers.

Discriminant: 4 + 4 = 8 → irrational roots.

So cannot factor further with integer coefficients.

Answer: z⁴ + 2z² – 1 (already factored as much as possible over integers)

Wait — maybe it’s a trick? Or perhaps typo?

Original says: “18 z⁴ + 2z² – 1”

No real factorization with integers. So leave as is.

But sometimes they expect to treat as quadratic in z² — but still doesn't factor.

So final: z⁴ + 2z² – 1

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Problem 19: h³ – k³

Difference of cubes: a³ – b³ = (a – b)(a² + ab + b²)

So: (h – k)(h² + hk + k²)

✔ Standard formula.

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Problem 20: 8p³ + r³

Sum of cubes: a³ + b³ = (a + b)(a² – ab + b²)

Here, 8p³ = (2p)³, r³ = r³

So: (2p + r)((2p)² – (2p)(r) + r²) = (2p + r)(4p² – 2pr + r²)

✔ Correct.

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Problem 21: 125h³ – 27k³

Difference of cubes: (5h)³ – (3k)³

= (5h – 3k)((5h)² + (5h)(3k) + (3k)²) = (5h – 3k)(25h² + 15hk + 9k²)

✔ Correct.

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Problem 22: 27c⁶ – 8d³

Note: 27c⁶ = (3c²)³, 8d³ = (2d)³

So difference of cubes: (3c² – 2d)((3c²)² + (3c²)(2d) + (2d)²) = (3c² – 2d)(9c⁴ + 6c²d + 4d²)

✔ Correct.

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Problem 23: 250a³ + 2b³

GCF: 2 → 2(125a³ + b³)

Now 125a³ = (5a)³, b³ = b³ → sum of cubes

= 2[(5a + b)((5a)² – (5a)(b) + b²)] = 2(5a + b)(25a² – 5ab + b²)

✔ Correct.

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Problem 24: 36w² – 49z²

Difference of squares: (6w)² – (7z)² = (6w – 7z)(6w + 7z)

✔ Simple.

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Problem 25: 5x² – 45

GCF: 5 → 5(x² – 9) = 5(x – 3)(x + 3)

✔ Difference of squares inside.

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Problem 26: a⁵x – 16ax³

GCF: ax → ax(a⁴ – 16x²)

Now a⁴ – 16x² = (a²)² – (4x)² = (a² – 4x)(a² + 4x)

And a² – 4x is not difference of squares unless 4x is square — but x is variable, so stop here.

Wait — a⁴ – 16x² = (a² – 4x)(a² + 4x) — correct.

But can we factor more? Only if 4x is a perfect square, which it’s not necessarily.

So: ax(a² – 4x)(a² + 4x)

But wait — original: a⁵x – 16a x³ = ax(a⁴ – 16x²) — yes.

Is a⁴ – 16x² factorable further? As above, yes into (a² – 4x)(a² + 4x)

So final: ax(a² – 4x)(a² + 4x)

✔ Check: ax(a²–4x)(a²+4x) = ax(a⁴ – 16x²) = a⁵x – 16a x³ ✔️

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Problem 27: 16a⁴ – b⁴

Difference of squares: (4a²)² – (b²)² = (4a² – b²)(4a² + b²)

Then 4a² – b² = (2a – b)(2a + b)

So: (2a – b)(2a + b)(4a² + b²)

✔ Fully factored.

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Problem 28: 3x⁷ + 36x⁵ + 108x³

GCF: 3x³ → 3x³(x⁴ + 12x² + 36)

Now x⁴ +12x² +36 = (x²)² + 2*6*x² + 6² = (x² + 6)²

So: 3x³(x² + 6)²

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

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Problem 29: 4u² + 20u + 25

Perfect square: (2u)² + 2*2u*5 + 5² = (2u + 5)²

✔ Yes.

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Problem 30: 121b² – 66bd + 9d²

Check: (11b)² = 121b², (3d)² = 9d², 2*11b*3d = 66bd → so (11b – 3d)²

Because middle term is negative.

So: (11b – 3d)²

✔ Check: (11b–3d)² = 121b² –66bd +9d² ✔️

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Now, compiling all answers:

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Final Answer:
1. (3p – 5)(p + 1)
2. 3(x – 5)(x + 3)
3. (4w + 5)(w – 5)
4. (2y – 5)²
5. 10j(j – 8)(j – 5)
6. (7 – 4b)(b – 7)
7. (2n + m)(n + 2m)
8. (4y – 1)(y – 4)
9. (3 – 5x)(x – 3)
10. 4(u + 1)(3u – 4)
11. (z² + 7)(z – 1)(z + 1)
12. 3(3k + 1)(k + 7)
13. (3x + y)(2x – 3y)
14. 3(2b + 1)(b – 3)
15. (4w + 3)(w – 5)
16. (4y – 5)(y – 3)
17. 3m(5m – 2)(m + 1)
18. z⁴ + 2z² – 1
19. (h – k)(h² + hk + k²)
20. (2p + r)(4p² – 2pr + r²)
21. (5h – 3k)(25h² + 15hk + 9k²)
22. (3c² – 2d)(9c⁴ + 6c²d + 4d²)
23. 2(5a + b)(25a² – 5ab + b²)
24. (6w – 7z)(6w + 7z)
25. 5(x – 3)(x + 3)
26. ax(a² – 4x)(a² + 4x)
27. (2a – b)(2a + b)(4a² + b²)
28. 3x³(x² + 6)²
29. (2u + 5)²
30. (11b – 3d)²