This algebra worksheet shows polynomial factoring problems with detailed work demonstrating grouping and special product techniques.

Let’s solve problem #9 step by step:

We are given:
(x³ + 3x²) – (5x – 15)

First, notice that we can factor each group separately.

Look at the first part: x³ + 3x²
Both terms have an x² in common.
So factor out x²:
→ x²(x + 3)

Now look at the second part: –(5x – 15)
This is the same as –5x + 15, but it’s better to keep the minus sign outside and factor inside.
Inside the parentheses: 5x – 15 → both terms have a 5 in common.
Factor out 5:
→ 5(x – 3)
But remember there’s a minus sign in front:
→ –5(x – 3)

Wait — let’s double-check that. Actually, if we factor 5x – 15, we get 5(x – 3).
But our original expression has –(5x – 15), which equals –5x + 15.
Alternatively, we can factor out –5 from –5x + 15:
→ –5(x – 3)? Let’s check: –5 × x = –5x, –5 × (–3) = +15 → yes! That works.

But wait — looking back at the student’s work in the image, they wrote:
x²(x + 3) – 5(x + 3)

That suggests they factored (5x – 15) as 5(x + 3)? That would be wrong because 5(x + 3) = 5x + 15, not 5x – 15.

Hold on — maybe there’s a typo in the problem or in their factoring?

Let me re-express the original problem carefully:

Original: (x³ + 3x²) – (5x – 15)

Distribute the negative sign:
= x³ + 3x² – 5x + 15

Now try grouping differently? Or perhaps the problem was meant to be:

(x³ + 3x²) – (5x + 15)? Then it would match the student’s work.

But as written, it’s (5x – 15).

Wait — let’s test the student’s answer:
They got: (x² – 5)(x + 3)

Multiply that back:
(x² – 5)(x + 3) = x²·x + x²·3 –5·x –5·3 = x³ + 3x² –5x –15

But our original expression after distributing is:
x³ + 3x² –5x +15

Ah! There’s a sign difference: we have +15, but the student’s factored form gives –15.

So either:
- The original problem has a typo, OR
- The student made a mistake in factoring.

Looking again at the problem as written in the image:
It says: (x³ + 3x²) – (5x – 15)

Which becomes: x³ + 3x² –5x +15

Can we factor this?

Try grouping:
Group as (x³ + 3x²) + (–5x + 15)

Factor each group:
x²(x + 3) –5(x – 3) ← note: –5x +15 = –5(x – 3)

Now we have:
x²(x + 3) –5(x – 3)

These do NOT have a common binomial factor. So we cannot factor further by grouping.

But the student wrote:
x²(x + 3) –5(x + 3) → which implies they thought –(5x – 15) = –5(x + 3), which is incorrect.

Actually, –(5x – 15) = –5x + 15 = –5(x – 3), not –5(x + 3).

So unless the problem was meant to be (x³ + 3x²) – (5x + 15), then the student’s answer would be correct.

Let me check what happens if the problem was:
(x³ + 3x²) – (5x + 15)

Then:
= x³ + 3x² –5x –15
Group: (x³ + 3x²) + (–5x –15)
= x²(x + 3) –5(x + 3)
= (x² – 5)(x + 3) ← matches student’s answer.

Given that the student’s work leads to a clean factored form, and the problem as written doesn’t factor nicely, it’s very likely that the intended problem was:

(x³ + 3x²) – (5x + 15)

In many textbooks, problems like this are designed to factor neatly by grouping.

Therefore, assuming the problem meant to say (5x + 15) instead of (5x – 15), here’s the correct solution:

Step 1: Write the expression:
(x³ + 3x²) – (5x + 15)

Step 2: Distribute the negative sign:
x³ + 3x² – 5x – 15

Step 3: Group terms:
(x³ + 3x²) + (–5x – 15)

Step 4: Factor each group:
x²(x + 3) –5(x + 3)

Step 5: Factor out the common binomial (x + 3):
(x + 3)(x² – 5)

Final Answer:
(x + 3)(x² – 5)

Note: If the problem truly is (5x – 15), then it does not factor nicely by grouping, and the expression remains as x³ + 3x² –5x +15, which doesn’t factor over integers. But since the student’s work suggests otherwise, and for educational purposes, we’ll go with the intended version.

Final Answer:
(x + 3)(x² – 5)