Matching binomials to their expanded forms in a math worksheet.

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Instructions:

Match the binomial to its expanded form.

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Problems:

1. (3x + 2y)⁵
2. (4x + 5y)²
3. (7x + 8y)⁴
4. (6x + 7y)³
5. (8x + 9y)²
6. (2x + 3y)⁴
7. (4x + 3y)³
8. (6x + 5y)²

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Options (a–h):

a. 216x³ + 756x²y + 882xy² + 343y³
b. 64x⁴ + 144x³y + 108x²y² + 27y⁴
c. x⁴ + 4xy + 6y²
d. 37x² + 54xy + 8y²
e. 81x⁴ + 216x³y + 216x²y² + 96xy³ + 16y⁴
f. 2401x⁴ + 2744x³y + 1176x²y² + 224xy³ + 16y⁴
g. 6x² + 24xy + 16y²
h. 16x² + 24xy + 9y²

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We need to match each binomial expression with its correct expansion using the Binomial Theorem:

> $(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Let’s go through each one step by step.

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1. (3x + 2y)⁵

Use binomial expansion:
$$
(3x + 2y)^5 = \sum_{k=0}^{5} \binom{5}{k} (3x)^{5-k} (2y)^k
$$

Compute terms:

- $k=0$: $\binom{5}{0}(3x)^5(2y)^0 = 1 \cdot 243x^5 = 243x^5$
- $k=1$: $\binom{5}{1}(3x)^4(2y)^1 = 5 \cdot 81x^4 \cdot 2y = 810x^4y$
- $k=2$: $\binom{5}{2}(3x)^3(2y)^2 = 10 \cdot 27x^3 \cdot 4y^2 = 1080x^3y^2$
- $k=3$: $\binom{5}{3}(3x)^2(2y)^3 = 10 \cdot 9x^2 \cdot 8y^3 = 720x^2y^3$
- $k=4$: $\binom{5}{4}(3x)^1(2y)^4 = 5 \cdot 3x \cdot 16y^4 = 240xy^4$
- $k=5$: $\binom{5}{5}(3x)^0(2y)^5 = 1 \cdot 32y^5 = 32y^5$

So:
$$
(3x + 2y)^5 = 243x^5 + 810x^4y + 1080x^3y^2 + 720x^2y^3 + 240xy^4 + 32y^5
$$

Now check if this matches any option → None of the options match this.

Wait — let's look again at the options. Option f:
> f. 2401x⁴ + 2744x³y + 1176x²y² + 224xy³ + 16y⁴

That has degree 4, but our expansion is degree 5 → not matching.

Hmm. Let's check other problems.

Maybe there's a typo or we’re missing something. Let’s try smaller ones first, which are easier.

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2. (4x + 5y)²

$$
(4x + 5y)^2 = (4x)^2 + 2(4x)(5y) + (5y)^2 = 16x^2 + 40xy + 25y^2
$$

Check options:

- h. 16x² + 24xy + 9y² → no
- g. 6x² + 24xy + 16y² → no
- d. 37x² + 54xy + 8y² → no
- None seem to match?

Wait — maybe I made a mistake.

Wait: (4x + 5y)² = 16x² + 40xy + 25y²

But none of the options have 40xy and 25y².

Wait — option h is: 16x² + 24xy + 9y² → that's (4x + 3y)²

Not matching.

Is it possible the options are for different binomials?

Wait — let’s look at option e:

> e. 81x⁴ + 216x³y + 216x²y² + 96xy³ + 16y⁴

This looks like (3x + 2y)⁴?

Let’s test: (3x + 2y)⁴

$$
(3x + 2y)^4 = \sum_{k=0}^{4} \binom{4}{k}(3x)^{4-k}(2y)^k
$$

- k=0: $\binom{4}{0}(3x)^4 = 81x^4$
- k=1: $\binom{4}{1}(3x)^3(2y) = 4 \cdot 27x^3 \cdot 2y = 216x^3y$
- k=2: $\binom{4}{2}(3x)^2(2y)^2 = 6 \cdot 9x^2 \cdot 4y^2 = 216x^2y^2$
- k=3: $\binom{4}{3}(3x)(2y)^3 = 4 \cdot 3x \cdot 8y^3 = 96xy^3$
- k=4: $\binom{4}{4}(2y)^4 = 16y^4$

Yes! So:
$$
(3x + 2y)^4 = 81x^4 + 216x^3y + 216x^2y^2 + 96xy^3 + 16y^4
$$

So option e matches (3x + 2y)⁴, but in the list, we have (3x + 2y)⁵ as #1.

Wait — so perhaps the worksheet has a typo? Or maybe we're supposed to find the best match.

Wait — let’s check #6: (2x + 3y)⁴

Try expanding (2x + 3y)⁴

$$
(2x + 3y)^4 = \sum_{k=0}^{4} \binom{4}{k}(2x)^{4-k}(3y)^k
$$

- k=0: $\binom{4}{0}(2x)^4 = 16x^4$
- k=1: $\binom{4}{1}(2x)^3(3y) = 4 \cdot 8x^3 \cdot 3y = 96x^3y$
- k=2: $\binom{4}{2}(2x)^2(3y)^2 = 6 \cdot 4x^2 \cdot 9y^2 = 216x^2y^2$
- k=3: $\binom{4}{3}(2x)(3y)^3 = 4 \cdot 2x \cdot 27y^3 = 216xy^3$
- k=4: $\binom{4}{4}(3y)^4 = 81y^4$

So:
$$
(2x + 3y)^4 = 16x^4 + 96x^3y + 216x^2y^2 + 216xy^3 + 81y^4
$$

Compare to option f:
> f. 2401x⁴ + 2744x³y + 1176x²y² + 224xy³ + 16y⁴

Nope — doesn’t match.

Wait — let’s try (7x + 8y)⁴ — #3

Try (7x + 8y)⁴

But that would be huge.

Wait — option f: 2401x⁴ + 2744x³y + 1176x²y² + 224xy³ + 16y⁴

Note: 2401 = 7⁴, 16 = 2⁴, but 8y⁴? No — wait: 2401 = 7⁴, 16 = 2⁴ → maybe (7x + 2y)⁴?

Try (7x + 2y)⁴

- k=0: $\binom{4}{0}(7x)^4 = 2401x^4$
- k=1: $\binom{4}{1}(7x)^3(2y) = 4 \cdot 343x^3 \cdot 2y = 2744x^3y$
- k=2: $\binom{4}{2}(7x)^2(2y)^2 = 6 \cdot 49x^2 \cdot 4y^2 = 1176x^2y^2$
- k=3: $\binom{4}{3}(7x)(2y)^3 = 4 \cdot 7x \cdot 8y^3 = 224xy^3$
- k=4: $\binom{4}{4}(2y)^4 = 16y^4$

Perfect! So:
$$
(7x + 2y)^4 = 2401x^4 + 2744x^3y + 1176x^2y^2 + 224xy^3 + 16y^4
$$

But in the list, we have (7x + 8y)⁴ — not (7x + 2y)⁴.

So unless there's a typo, it doesn’t match.

Wait — perhaps option f is for (7x + 2y)⁴, but we don’t have that binomial.

Let’s re-express the binomials and see which expansions match.

Let’s go back to #5: (8x + 9y)²

$$
(8x + 9y)^2 = 64x^2 + 2\cdot8x\cdot9y + 81y^2 = 64x^2 + 144xy + 81y^2
$$

Check options:

- b: 64x⁴ + 144x³y + 108x²y² + 27y⁴ → degree 4, no
- g: 6x² + 24xy + 16y² → no
- h: 16x² + 24xy + 9y² → no

Wait — none match.

Wait — option b is: 64x⁴ + 144x³y + 108x²y² + 27y⁴

Try (4x + 3y)³?

(4x + 3y)³ = ?

- (4x)^3 = 64x³
- 3(4x)²(3y) = 3*16x²*3y = 144x²y
- 3(4x)(3y)² = 3*4x*9y² = 108xy²
- (3y)³ = 27y³

So: 64x³ + 144x²y + 108xy² + 27y³

But option b is: 64x⁴ + 144x³y + 108x²y² + 27y⁴ → degree 4

So maybe (4x + 3y)⁴?

Try (4x + 3y)⁴

- k=0: $\binom{4}{0}(4x)^4 = 256x^4$
- k=1: $\binom{4}{1}(4x)^3(3y) = 4 * 64x^3 * 3y = 768x^3y$
- Not matching.

Wait — option b: 64x⁴ + 144x³y + 108x²y² + 27y⁴

Coefficients: 64, 144, 108, 27

Note: 64 = 4⁴, 27 = 3⁴ → maybe (4x + 3y)⁴?

But earlier calculation gave 256x⁴, not 64.

Wait — 64 = 2⁶? No.

Wait — 64 = 8²? Hmm.

Wait — could it be (2x + 3y)⁴?

Earlier: (2x + 3y)⁴ = 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴ → no

Wait — option a: 216x³ + 756x²y + 882xy² + 343y³

343 = 7³ → maybe (6x + 7y)³?

Try (6x + 7y)³

- (6x)^3 = 216x³
- 3(6x)²(7y) = 3*36x²*7y = 756x²y
- 3(6x)(7y)² = 3*6x*49y² = 882xy²
- (7y)³ = 343y³

Perfect!

So:
$$
(6x + 7y)^3 = 216x^3 + 756x^2y + 882xy^2 + 343y^3
$$

And this is option a

So #4: (6x + 7y)³ → a

Great!

Now #1: (3x + 2y)⁵ — we computed earlier, but it’s degree 5, and all options are degree ≤ 4.

But option e is degree 4: (3x + 2y)⁴

So likely, #1 should be (3x + 2y)⁴, but it's written as ^5.

Wait — maybe typo in the worksheet?

Alternatively, let’s look at #3: (7x + 8y)⁴

Try to expand (7x + 8y)⁴ — too big.

But option f is: 2401x⁴ + 2744x³y + 1176x²y² + 224xy³ + 16y⁴

We already saw that this is (7x + 2y)⁴

But we have (7x + 8y)⁴ — not matching.

Wait — what about (8x + 9y)²?

We had: 64x² + 144xy + 81y²

But no option has that.

Wait — option g: 6x² + 24xy + 16y² → (2x + 4y)² = 4x² + 16xy + 16y² → no

Wait — (3x + 4y)² = 9x² + 24xy + 16y² → close to g

But g is 6x² + 24xy + 16y² → not matching.

Wait — option h: 16x² + 24xy + 9y² → (4x + 3y)² = 16x² + 24xy + 9y² → YES!

So (4x + 3y)² → h

But in the list, we have #2: (4x + 5y)² → not matching.

Wait — maybe #2 is (4x + 3y)²?

But it says (4x + 5y)²

(4x + 5y)² = 16x² + 40xy + 25y² → not in options.

Wait — option d: 37x² + 54xy + 8y²

Does that match any?

Try (6x + y)² = 36x² + 12xy + y² → no

(5x + 3y)² = 25x² + 30xy + 9y² → no

(7x + 2y)² = 49x² + 28xy + 4y² → no

(8x + y)² = 64x² + 16xy + y² → no

Not matching.

Wait — maybe #5: (8x + 9y)² = 64x² + 144xy + 81y²

Is there an option like that?

No.

Wait — option c: x⁴ + 4xy + 6y² — seems wrong; degrees don’t match.

Wait — c is x⁴ + 4xy + 6y² — that can't be right.

Probably typo.

Wait — perhaps c is meant to be (x + y)² but it's written as x⁴...

No — it says x⁴ + 4xy + 6y² — impossible.

Wait — maybe it's (x + y)⁴?

(x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

But c is x⁴ + 4xy + 6y² — not matching.

So likely c is a typo.

Wait — let’s try #7: (4x + 3y)³

(4x + 3y)³ = ?

- (4x)^3 = 64x³
- 3*(4x)²*(3y) = 3*16x²*3y = 144x²y
- 3*(4x)*(3y)² = 3*4x*9y² = 108xy²
- (3y)³ = 27y³

So: 64x³ + 144x²y + 108xy² + 27y³

But option b is: 64x⁴ + 144x³y + 108x²y² + 27y⁴ — degree 4

So if it were (4x + 3y)⁴, then:

(4x + 3y)⁴ = ?

- k=0: $\binom{4}{0}(4x)^4 = 256x^4$
- k=1: $\binom{4}{1}(4x)^3(3y) = 4*64x^3*3y = 768x^3y$
- k=2: $\binom{4}{2}(4x)^2(3y)^2 = 6*16x^2*9y^2 = 864x^2y^2$
- k=3: $\binom{4}{3}(4x)(3y)^3 = 4*4x*27y^3 = 432xy^3$
- k=4: $\binom{4}{4}(3y)^4 = 81y^4$

Not matching option b.

But option b is: 64x⁴ + 144x³y + 108x²y² + 27y⁴

64 = 4⁴? 4⁴ = 256, no.

Wait — 64 = 8²? 64 = 2⁶?

Wait — 64 = 4³? No.

Wait — could it be (2x + 3y)⁴?

We did: 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴ — no.

Wait — option a we matched to (6x + 7y)³

So #4 → a

Now #6: (2x + 3y)⁴ — we computed: 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴

Is there an option like that?

No.

But option e is: 81x⁴ + 216x³y + 216x²y² + 96xy³ + 16y⁴

This is very close to (2x + 3y)⁴, but reversed: 81x⁴ vs 16x⁴

So this is (3x + 2y)⁴

Yes! As we saw earlier.

So (3x + 2y)⁴ → e

But in the list, #1 is (3x + 2y)⁵ — probably a typo, should be ^4

Similarly, #3: (7x + 8y)⁴ — but we have (7x + 2y)⁴ matching f

So maybe typo in binomial.

Wait — option f: 2401x⁴ + 2744x³y + 1176x²y² + 224xy³ + 16y⁴

We said this is (7x + 2y)⁴

But we have (7x + 8y)⁴ — not matching.

Unless it's (7x + 2y)⁴, but listed as (7x + 8y)⁴ — typo.

Alternatively, maybe the binomial is (7x + 2y)⁴, but written as (7x + 8y)⁴.

Likely typos.

Let’s assume the worksheet has some typos, and match based on expansions.

Let’s list what we know:

- (6x + 7y)³ → a → #4
- (3x + 2y)⁴ → e → should be #1, but it's written as ^5
- (7x + 2y)⁴ → f → should be #3, but it's (7x + 8y)⁴
- (4x + 3y)² → h → should be #2, but it's (4x + 5y)²
- (8x + 9y)² → ? → 64x² + 144xy + 81y² — not in options
- (2x + 3y)⁴ → 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴ — not in options
- (4x + 3y)³ → 64x³ + 144x²y + 108xy² + 27y³ — not in options
- (6x + 5y)² → 36x² + 60xy + 25y² — not in options

Wait — option b: 64x⁴ + 144x³y + 108x²y² + 27y⁴

This is degree 4, coefficients: 64, 144, 108, 27

64 = 4⁴? 4⁴=256, no.

64 = 8², 27 = 3³

Wait — 64 = 4³? 4³=64, yes.

Wait — try (4x + 3y)⁴? We did: 256x⁴, no.

Wait — (2x + 3y)⁴ = 16x⁴ + 96x³y + 216x²y² + 216xy³ + 81y⁴

No.

Wait — (8x + 3y)² = 64x² + 48xy + 9y² — no.

Wait — option b is 64x⁴ + 144x³y + 108x²y² + 27y⁴

Try (4x + 3y)⁴? No.

Wait — (8x + 3y)⁴? Too big.

Perhaps it's (4x + 3y)⁴ with typo.

Wait — maybe it's (4x + 3y)³, but degree 3, but it's degree 4.

No.

Another idea: maybe the binomial is (2x + 3y)⁴, but written as (2x + 3y)³.

But no.

Let’s look at #8: (6x + 5y)²

(6x + 5y)² = 36x² + 60xy + 25y²

Not in options.

Option g: 6x² + 24xy + 16y²

Try (x + 4y)² = x² + 8xy + 16y² — no

(2x + 4y)² = 4x² + 16xy + 16y² — no

(3x + 4y)² = 9x² + 24xy + 16y² — close to g

g is 6x² + 24xy + 16y² — not matching.

(√6x + 4y)² — not integer.

So likely, the only matches are:

- (6x + 7y)³ → a → #4
- (3x + 2y)⁴ → e → #1 (but should be ^4)
- (7x + 2y)⁴ → f → #3 (should be (7x + 2y)⁴)
- (4x + 3y)² → h → #2 (should be (4x + 3y)²)

But in the list, #2 is (4x + 5y)², #1 is (3x + 2y)⁵, etc.

So likely, there are typos in the binomials.

Alternatively, maybe the options are correct, and we need to find which binomial matches.

Let’s reverse-engineer.

Look at option b: 64x⁴ + 144x³y + 108x²y² + 27y⁴

Assume it's (ax + by)⁴

Then:
- First term: a⁴x⁴ = 64x⁴ → a⁴ = 64 → a = ∛64? 64 = 2⁶, so a = 2^(3/2)? No.

64 = 4³, but 4³ = 64, not a⁴.

64 = 2⁶, so a⁴ = 2⁶ → a = 2^(3/2) = 2√2 — not integer.

But last term: b⁴y⁴ = 27y⁴ → b⁴ = 27 → b = 27^(1/4) — not nice.

So unlikely.

Wait — option d: 37x² + 54xy + 8y²

Try to factor: discriminant = 54² - 4*37*8 = 2916 - 1184 = 1732 — not perfect square.

So not a perfect square.

Wait — maybe it's (6x + y)² = 36x² + 12xy + y² — no

(7x + 2y)² = 49x² + 28xy + 4y² — no

(8x + y)² = 64x² + 16xy + y² — no

(5x + 3y)² = 25x² + 30xy + 9y² — no

(6x + 3y)² = 36x² + 36xy + 9y² — no

No match.

Perhaps the only correct matches are:

- #4: (6x + 7y)³ → a
- #1: (3x + 2y)⁵ — but no match, but (3x + 2y)⁴ → e
- #3: (7x + 8y)⁴ — no match, but (7x + 2y)⁴ → f
- #2: (4x + 5y)² — no match, but (4x + 3y)² → h
- #5: (8x + 9y)² — no match
- #6: (2x + 3y)⁴ — no match, but (3x + 2y)⁴ → e
- #7: (4x + 3y)³ — no match
- #8: (6x + 5y)² — no match

So likely, the worksheet has errors in the binomials.

Perhaps the intended matches are:

1. (3x + 2y)⁴ → e
2. (4x + 3y)² → h
3. (7x + 2y)⁴ → f
4. (6x + 7y)³ → a
5. (8x + 9y)² → ? (not in options)
6. (2x + 3y)⁴ → ? (not in options)
7. (4x + 3y)³ → ? (not in options)
8. (6x + 5y)² → ? (not in options)

But since only a, e, f, h are plausible, and we have four of them, perhaps only those are intended.

So based on the most likely matches:

- (6x + 7y)³ → a
- (3x + 2y)⁴ → e
- (7x + 2y)⁴ → f
- (4x + 3y)² → h

But since the worksheet lists different binomials, the only one that matches exactly is:

✔ #4: (6x + 7y)³ → a

All others are likely typos.

If you can provide the image or confirm the exact binomials, I can give a more accurate answer.

But based on the given data, the only clear match is:

> 4. (6x + 7y)³ → a

Other matches require assuming typos in the binomials.

For example:

- If #1 was (3x + 2y)⁴, then → e
- If #2 was (4x + 3y)², then → h
- If #3 was (7x + 2y)⁴, then → f

Otherwise, no other matches.

So final answer (assuming typos):

| Binomial | Match |
|---------|-------|
| 1. (3x + 2y)⁵ | (likely typo, should be ^4) → e |
| 2. (4x + 5y)² | (likely typo, should be (4x + 3y)²) → h |
| 3. (7x + 8y)⁴ | (likely typo, should be (7x + 2y)⁴) → f |
| 4. (6x + 7y)³ | → a |
| 5. (8x + 9y)² | no match |
| 6. (2x + 3y)⁴ | no match |
| 7. (4x + 3y)³ | no match |
| 8. (6x + 5y)² | no match |

But since only a, e, f, h are used, and only a is perfectly matched to a given binomial, the only certain answer is:

> 4. (6x + 7y)³ → a

Please double-check the worksheet for typos.

If you can confirm the binomials or the options, I can refine this.