Factoring Polynomials: Cubic Polynomials worksheet with 11 algebraic expressions to factor.

Let's solve the problem step by step.

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1) Write the pattern for the sum of two cubes:

The sum of two cubes follows this algebraic identity:

$$
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
$$

This is the standard factoring pattern for the sum of two cubes.

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Now, let's factor each expression using this pattern or other appropriate methods.

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2) $ x^3 + 64 $

Note: $ 64 = 4^3 $

So:
$$
x^3 + 64 = x^3 + 4^3 = (x + 4)(x^2 - 4x + 16)
$$

✔ Answer: $ (x + 4)(x^2 - 4x + 16) $

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3) $ x^3 + 343 $

$ 343 = 7^3 $

So:
$$
x^3 + 343 = x^3 + 7^3 = (x + 7)(x^2 - 7x + 49)
$$

✔ Answer: $ (x + 7)(x^2 - 7x + 49) $

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4) $ x^3 + 125 $

$ 125 = 5^3 $

$$
x^3 + 125 = x^3 + 5^3 = (x + 5)(x^2 - 5x + 25)
$$

✔ Answer: $ (x + 5)(x^2 - 5x + 25) $

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5) $ 2x^3 + 54 $

First, factor out the GCF: $ 2 $

$$
2x^3 + 54 = 2(x^3 + 27)
$$

Now, $ 27 = 3^3 $, so:

$$
x^3 + 27 = x^3 + 3^3 = (x + 3)(x^2 - 3x + 9)
$$

So:
$$
2x^3 + 54 = 2(x + 3)(x^2 - 3x + 9)
$$

✔ Answer: $ 2(x + 3)(x^2 - 3x + 9) $

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6) $ ab^4c^4 + abc^4 $

Let’s look at this carefully.

We have:
$$
ab^4c^4 + abc^4
$$

Factor out the greatest common factor (GCF).

Both terms have:
- $ a $
- $ b $
- $ c^4 $

So:
$$
= abc^4(b^3 + 1)
$$

Now, notice that $ b^3 + 1 $ is a sum of cubes:
$$
b^3 + 1 = b^3 + 1^3 = (b + 1)(b^2 - b + 1)
$$

So:
$$
ab^4c^4 + abc^4 = abc^4(b + 1)(b^2 - b + 1)
$$

✔ Answer: $ abc^4(b + 1)(b^2 - b + 1) $

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7) $ x^3 + 216 $

$ 216 = 6^3 $

So:
$$
x^3 + 216 = x^3 + 6^3 = (x + 6)(x^2 - 6x + 36)
$$

✔ Answer: $ (x + 6)(x^2 - 6x + 36) $

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8) $ x^3 + 512 $

$ 512 = 8^3 $

$$
x^3 + 512 = x^3 + 8^3 = (x + 8)(x^2 - 8x + 64)
$$

✔ Answer: $ (x + 8)(x^2 - 8x + 64) $

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9) $ 3x^3y^2 + 24y^2 $

Factor out the GCF: both terms have $ 3y^2 $

$$
= 3y^2(x^3 + 8)
$$

Now, $ x^3 + 8 = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4) $

So:
$$
3x^3y^2 + 24y^2 = 3y^2(x + 2)(x^2 - 2x + 4)
$$

✔ Answer: $ 3y^2(x + 2)(x^2 - 2x + 4) $

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10) $ 2x^4 + 250x $

Factor out GCF: $ 2x $

$$
= 2x(x^3 + 125)
$$

Now, $ 125 = 5^3 $, so:
$$
x^3 + 125 = x^3 + 5^3 = (x + 5)(x^2 - 5x + 25)
$$

So:
$$
2x^4 + 250x = 2x(x + 5)(x^2 - 5x + 25)
$$

✔ Answer: $ 2x(x + 5)(x^2 - 5x + 25) $

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11) $ 5x^4 + 320 $

Factor out GCF: $ 5 $

$$
= 5(x^4 + 64)
$$

Wait — $ x^4 + 64 $ is not a sum of cubes. But note: $ x^4 = (x^2)^2 $, and $ 64 = 8^2 $, so this is a sum of squares, which does not factor over real numbers using standard factoring patterns.

But wait — can we write it as a sum of cubes? Let's check:

Is $ x^4 + 64 $ a sum of cubes?

No — because $ x^4 $ is not a cube unless we rewrite it.

But $ x^4 = (x^{4/3})^3 $? Not helpful.

Alternatively, perhaps we made a mistake in interpretation.

Wait — $ 5x^4 + 320 = 5(x^4 + 64) $

But $ x^4 + 64 $ is not a sum of cubes.

But note: $ x^4 + 64 $ can be rewritten as $ x^4 + 16x^2 + 64 - 16x^2 = (x^2 + 8)^2 - (4x)^2 $, which is a difference of squares, but that's not required here.

Wait — is there a typo? Or are we missing something?

Let’s double-check: Is $ x^4 + 64 $ factorable?

Actually, $ x^4 + 64 $ can be factored using sum of squares trick:

$$
x^4 + 64 = x^4 + 16x^2 + 64 - 16x^2 = (x^2 + 8)^2 - (4x)^2 = (x^2 + 8 - 4x)(x^2 + 8 + 4x)
$$

So:
$$
x^4 + 64 = (x^2 - 4x + 8)(x^2 + 4x + 8)
$$

Therefore:
$$
5x^4 + 320 = 5(x^4 + 64) = 5(x^2 - 4x + 8)(x^2 + 4x + 8)
$$

But this is not a sum of cubes — it's a sum of fourth powers, and we used difference of squares technique.

However, the original problem says "Factoring Polynomials: Cubic Polynomials", so maybe this one is not cubic?

Wait — $ 5x^4 + 320 $ is degree 4, not cubic.

So perhaps it's a mistake, or it's included to test recognition.

But let's see: is there any way to interpret this as a sum of cubes?

Try: $ x^4 + 64 = (x^{4/3})^3 + 4^3 $? No, not valid.

Alternatively, perhaps it was meant to be $ x^3 + 64 $? But no, it says $ 5x^4 + 320 $.

Wait — what if we factor $ 5x^4 + 320 $ differently?

$$
5x^4 + 320 = 5(x^4 + 64)
$$

And as shown above, $ x^4 + 64 $ factors into quadratics.

So even though it's not a cubic, it can be factored.

But since the worksheet is about cubic polynomials, this might be an error.

Alternatively, perhaps it's supposed to be $ 5x^3 + 320 $? Let’s check:

If $ 5x^3 + 320 $, then factor out 5:
$$
5(x^3 + 64) = 5(x + 4)(x^2 - 4x + 16)
$$

That would make sense.

But as written: $ 5x^4 + 320 $

So unless it's a typo, we proceed with:

$$
5x^4 + 320 = 5(x^4 + 64) = 5(x^2 - 4x + 8)(x^2 + 4x + 8)
$$

✔ Answer: $ 5(x^2 - 4x + 8)(x^2 + 4x + 8) $

(Note: This is not a sum of cubes, but it is fully factored.)

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✔ Summary of Answers:

1) Sum of two cubes pattern:
$$
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
$$

2) $ x^3 + 64 = (x + 4)(x^2 - 4x + 16) $

3) $ x^3 + 343 = (x + 7)(x^2 - 7x + 49) $

4) $ x^3 + 125 = (x + 5)(x^2 - 5x + 25) $

5) $ 2x^3 + 54 = 2(x + 3)(x^2 - 3x + 9) $

6) $ ab^4c^4 + abc^4 = abc^4(b + 1)(b^2 - b + 1) $

7) $ x^3 + 216 = (x + 6)(x^2 - 6x + 36) $

8) $ x^3 + 512 = (x + 8)(x^2 - 8x + 64) $

9) $ 3x^3y^2 + 24y^2 = 3y^2(x + 2)(x^2 - 2x + 4) $

10) $ 2x^4 + 250x = 2x(x + 5)(x^2 - 5x + 25) $

11) $ 5x^4 + 320 = 5(x^2 - 4x + 8)(x^2 + 4x + 8) $

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Final Notes:

- Most problems use the sum of cubes formula.
- Problem 11 is not a cubic polynomial — it's degree 4 — so likely either a typo or extra challenge.
- Always factor out constants/GCF first when possible.
- Remember: $ a^3 + b^3 = (a + b)(a^2 - ab + b^2) $

Let me know if you'd like these explained further!