Factoring Polynomials worksheet with 12 algebraic expressions to solve.

Let’s solve each polynomial factoring problem one by one. I’ll explain the method used for each.

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Problem 1: \(3x^2 - 2x - 5\)

We factor using the AC method (for trinomials of form \(ax^2 + bx + c\)).

- Multiply \(a \cdot c = 3 \cdot (-5) = -15\)
- Find two numbers that multiply to -15 and add to -2 → 3 and -5
- Rewrite middle term:
\(3x^2 + 3x - 5x - 5\)
- Group:
\((3x^2 + 3x) + (-5x - 5)\)
- Factor each group:
\(3x(x + 1) -5(x + 1)\)
- Factor out common binomial:
\((3x - 5)(x + 1)\)

✔ Answer: \((3x - 5)(x + 1)\)

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Problem 2: \(10m^2 + 17m + 6\)

AC method again:

- \(a \cdot c = 10 \cdot 6 = 60\)
- Numbers that multiply to 60 and add to 17 → 12 and 5
- Rewrite:
\(10m^2 + 12m + 5m + 6\)
- Group:
\((10m^2 + 12m) + (5m + 6)\)
- Factor:
\(2m(5m + 6) + 1(5m + 6)\)
- Common binomial:
\((2m + 1)(5m + 6)\)

✔ Answer: \((2m + 1)(5m + 6)\)

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Problem 3: \(2m^2 + 63m + 145\)

AC method:

- \(a \cdot c = 2 \cdot 145 = 290\)
- Need two numbers that multiply to 290 and add to 63 → 58 and 5
- Rewrite:
\(2m^2 + 58m + 5m + 145\)
- Group:
\((2m^2 + 58m) + (5m + 145)\)
- Factor:
\(2m(m + 29) + 5(m + 29)\)
- Common binomial:
\((2m + 5)(m + 29)\)

✔ Answer: \((2m + 5)(m + 29)\)

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Problem 4: \(3x^7 + 36x^5 + 108x^3\)

First, factor out the GCF (greatest common factor):

- GCF is \(3x^3\):
\(3x^3(x^4 + 12x^2 + 36)\)
- Now factor the trinomial inside:
\(x^4 + 12x^2 + 36 = (x^2 + 6)^2\) ← perfect square trinomial

✔ Answer: \(3x^3(x^2 + 6)^2\)

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Problem 5: \(12v^2 - 4v - 16\)

First, factor out GCF: all terms divisible by 4:

- \(4(3v^2 - v - 4)\)
- Now factor \(3v^2 - v - 4\)

AC method:

- \(a \cdot c = 3 \cdot (-4) = -12\)
- Numbers: 3 and -4 → sum = -1? Wait — 3 + (-4) = -1 → yes!
- Rewrite:
\(3v^2 + 3v - 4v - 4\)
- Group:
\((3v^2 + 3v) + (-4v - 4)\)
- Factor:
\(3v(v + 1) -4(v + 1)\)
- Common binomial:
\((3v - 4)(v + 1)\)

✔ Answer: \(4(3v - 4)(v + 1)\)

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Problem 6: \(3x^2 - 8x + 4\)

AC method:

- \(a \cdot c = 3 \cdot 4 = 12\)
- Numbers that multiply to 12 and add to -8 → -6 and -2
- Rewrite:
\(3x^2 - 6x - 2x + 4\)
- Group:
\((3x^2 - 6x) + (-2x + 4)\)
- Factor:
\(3x(x - 2) -2(x - 2)\)
- Common binomial:
\((3x - 2)(x - 2)\)

✔ Answer: \((3x - 2)(x - 2)\)

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Problem 7: \(10u^3 - 130u^2 + 400u\)

Factor out GCF: all terms divisible by 10u:

- \(10u(u^2 - 13u + 40)\)
- Factor quadratic: find two numbers multiplying to 40, adding to -13 → -8 and -5

✔ Answer: \(10u(u - 8)(u - 5)\)

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Problem 8: \(121a^2 - 66ab + 9b^2\)

This looks like a perfect square trinomial:

Check:
- First term: \((11a)^2 = 121a^2\)
- Last term: \((3b)^2 = 9b^2\)
- Middle term: \(2 \cdot 11a \cdot 3b = 66ab\) → matches with negative sign

So:
\((11a - 3b)^2\)

✔ Answer: \((11a - 3b)^2\)

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Problem 9: \(4w^2 - 17w - 15\)

AC method:

- \(a \cdot c = 4 \cdot (-15) = -60\)
- Numbers: -20 and 3 → sum = -17
- Rewrite:
\(4w^2 - 20w + 3w - 15\)
- Group:
\((4w^2 - 20w) + (3w - 15)\)
- Factor:
\(4w(w - 5) + 3(w - 5)\)
- Common binomial:
\((4w + 3)(w - 5)\)

✔ Answer: \((4w + 3)(w - 5)\)

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Problem 10: \(2k^2 + 19k - 10\)

AC method:

- \(a \cdot c = 2 \cdot (-10) = -20\)
- Numbers: 20 and -1 → sum = 19
- Rewrite:
\(2k^2 + 20k - k - 10\)
- Group:
\((2k^2 + 20k) + (-k - 10)\)
- Factor:
\(2k(k + 10) -1(k + 10)\)
- Common binomial:
\((2k - 1)(k + 10)\)

✔ Answer: \((2k - 1)(k + 10)\)

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Problem 11: \(2m^2 + 63m + 145\)

Wait — this is identical to Problem 3!

✔ Answer: \((2m + 5)(m + 29)\)

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Problem 12: \(15a^2 + 45ab + 60ab^2\)

First, check for GCF:

- All terms divisible by 15a? Let's see:
- \(15a^2\), \(45ab\), \(60ab^2\) → GCF is 15a

Factor out 15a:

- \(15a(a + 3b + 4b^2)\)

Wait — inside: \(a + 3b + 4b^2\) — this is not a standard polynomial in one variable. It has mixed variables.

Actually, let me re-express:

The expression is:
\(15a^2 + 45ab + 60ab^2\)

Group terms with common factors:

But notice — the last term is \(60ab^2\), which is different from \(60a^2b\) or something else.

Actually, perhaps it’s meant to be \(15a^2 + 45ab + 60b^2\)? But as written, it’s \(60ab^2\).

Let’s factor as is:

Factor out 15a:

→ \(15a(a + 3b + 4b^2)\)

Now, can we factor \(a + 3b + 4b^2\)? Not really — it’s not factorable over integers because it’s linear in ‘a’ and quadratic in ‘b’. So unless there’s a typo, this is as far as we can go.

But wait — maybe the problem is miswritten? Let me double-check.

If it were \(15a^2 + 45ab + 60b^2\), then:

- GCF = 15 → \(15(a^2 + 3ab + 4b^2)\) → but \(a^2 + 3ab + 4b^2\) doesn’t factor nicely.

Alternatively, if it were \(15a^2 + 45ab + 60a^2b^2\)? No.

Given as is: \(15a^2 + 45ab + 60ab^2\)

Factor out 15a:

✔ Answer: \(15a(a + 3b + 4b^2)\)

*(Note: This may be a typo in the original worksheet — often such problems are meant to be quadratics in two variables with symmetric structure. But based on what’s written, this is correct.)*

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## ✔ Final Answers Summary:

1. \((3x - 5)(x + 1)\)
2. \((2m + 1)(5m + 6)\)
3. \((2m + 5)(m + 29)\)
4. \(3x^3(x^2 + 6)^2\)
5. \(4(3v - 4)(v + 1)\)
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. \((2m + 5)(m + 29)\) *(same as #3)*
12. \(15a(a + 3b + 4b^2)\)

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Let me know if you’d like to see step-by-step work for any specific problem!