Factoring difference of two squares online exercise for | Live ... - Free Printable

Let’s solve each problem step by step. All of these are differences of squares, which follow this pattern:

> a² – b² = (a – b)(a + b)

We’ll apply that to each one.

---

1. y² – 1

- y² is a square → (y)²
- 1 is a square → (1)²
- So:
y² – 1 = (y – 1)(y + 1)

---

2. 9u² – 4

- 9u² = (3u)²
- 4 = (2)²
- So:
9u² – 4 = (3u – 2)(3u + 2)

---

3. 144x² – 25

- 144x² = (12x)²
- 25 = (5)²
- So:
144x² – 25 = (12x – 5)(12x + 5)

---

4. 121 – 9y²

- 121 = (11)²
- 9y² = (3y)²
- So:
121 – 9y² = (11 – 3y)(11 + 3y)
*(You can also write it as (–3y + 11)(3y + 11), but usually we keep the positive term first if possible — either way is correct, but let’s stick with order given: constant minus variable term)*

Actually, since subtraction is not commutative, better to write as:
(11 – 3y)(11 + 3y) — this matches the original order.

---

5. 81x² – 121

- 81x² = (9x)²
- 121 = (11)²
- So:
81x² – 121 = (9x – 11)(9x + 11)

---

6. 121y² – 36

- 121y² = (11y)²
- 36 = (6)²
- So:
121y² – 36 = (11y – 6)(11y + 6)

---

7. 64a² – 25

- 64a² = (8a)²
- 25 = (5)²
- So:
64a² – 25 = (8a – 5)(8a + 5)

---

8. 25u² – 1

- 25u² = (5u)²
- 1 = (1)²
- So:
25u² – 1 = (5u – 1)(5u + 1)

---

9. 49x² – 4

- 49x² = (7x)²
- 4 = (2)²
- So:
49x² – 4 = (7x – 2)(7x + 2)

---

10. 81 – 36x²

First, notice both terms have a common factor? Let’s check:

81 and 36 are both divisible by 9.

So factor out 9 first:

→ 9(9 – 4x²)

Now look inside: 9 – 4x² → difference of squares!

- 9 = (3)²
- 4x² = (2x)²

So:
9 – 4x² = (3 – 2x)(3 + 2x)

Therefore, full factoring:

81 – 36x² = 9(3 – 2x)(3 + 2x)

*(Note: You could also write it as 9(–2x + 3)(2x + 3), but again, keeping positive leading terms in parentheses is cleaner. Also, some might prefer to factor out -9 to make the x-term positive, but unless specified, factoring out the GCF and then applying difference of squares is standard.)*

Wait — actually, let me double-check: Is there a reason to factor further? No, because 9 is already factored, and (3 – 2x)(3 + 2x) is fully factored.

But note: Sometimes people like to write the variable term first. However, mathematically, (3 – 2x) is fine. Alternatively, you can factor out a negative from one binomial to rearrange, but that’s unnecessary unless instructed.

Alternatively, we could write:

81 – 36x² = –(36x² – 81) = –[ (6x)² – 9² ] = –(6x – 9)(6x + 9)

Then factor out 3 from each binomial:

= –[3(2x – 3)][3(2x + 3)] = –9(2x – 3)(2x + 3)

Which equals 9(3 – 2x)(3 + 2x) — same thing.

So both forms are equivalent. But since the question says “factor completely”, and we’ve pulled out the GCF and applied difference of squares, 9(3 – 2x)(3 + 2x) is acceptable.

However, sometimes teachers prefer the variable term to come first in the binomials. In that case, we can write:

9(–2x + 3)(2x + 3) — still messy.

Or better: Factor out –9 initially?

Let’s try:

81 – 36x² = –36x² + 81 = –9(4x² – 9) = –9[(2x)² – 3²] = –9(2x – 3)(2x + 3)

This gives us: –9(2x – 3)(2x + 3)

But now we have a negative sign. Is that “completely factored”? Yes, but perhaps less preferred.

Actually, the most standard approach is to factor out the greatest common factor *without* introducing negatives unless necessary.

Original expression: 81 – 36x²

GCF of 81 and 36 is 9.

So: 9(9 – 4x²) = 9(3² – (2x)²) = 9(3 – 2x)(3 + 2x)

✔ This is correct and complete.

Some may argue that (3 – 2x) should be written as –(2x – 3), so:

9 × [–(2x – 3)] × (3 + 2x) = –9(2x – 3)(2x + 3)

But again, unless specified, both are correct. However, in many textbooks, they leave it as 9(3 – 2x)(3 + 2x).

To match typical expectations, I’ll go with:

9(3 – 2x)(3 + 2x)

But wait — let me verify numerically.

Pick x = 1:

Original: 81 – 36(1)² = 81 – 36 = 45

Factored: 9(3 – 2)(3 + 2) = 9(1)(5) = 45 ✔

If I used –9(2x – 3)(2x + 3):

At x=1: –9(2–3)(2+3) = –9(-1)(5) = 45 ✔

Same result.

But since the problem didn’t specify form, and 9(3 – 2x)(3 + 2x) avoids leading negatives, I’ll use that.

However, upon second thought — in algebra, it's often preferred to have the variable term first in the binomial when possible. For example, (2x – 3) instead of (3 – 2x). So maybe we should adjust.

Let’s do:

9(3 – 2x)(3 + 2x) = 9 × [–1(2x – 3)] × (2x + 3) = –9(2x – 3)(2x + 3)

But now we have a negative coefficient. Is that worse? Not really — it’s still factored completely.

Actually, looking at other problems, none had negative signs introduced, so perhaps for consistency, we should avoid introducing a negative unless required.

In problem #4: 121 – 9y² = (11 – 3y)(11 + 3y) — here, the variable term is second, and no negative was introduced.

Similarly, here, 81 – 36x² = 9(9 – 4x²) = 9(3 – 2x)(3 + 2x) — consistent with #4.

So I think it’s fine.

Final decision: 9(3 – 2x)(3 + 2x)

But let me check online or standard practice... Actually, in many sources, they would factor out the GCF and then write the difference of squares as is, even if the variable comes second.

Example: Khan Academy often leaves it as (a – b)(a + b) regardless of variable position.

So I’ll stick with:

9(3 – 2x)(3 + 2x)

---

Now, compiling all answers:

Final Answer:
1. (y – 1)(y + 1)
2. (3u – 2)(3u + 2)
3. (12x – 5)(12x + 5)
4. (11 – 3y)(11 + 3y)
5. (9x – 11)(9x + 11)
6. (11y – 6)(11y + 6)
7. (8a – 5)(8a + 5)
8. (5u – 1)(5u + 1)
9. (7x – 2)(7x + 2)
10. 9(3 – 2x)(3 + 2x)