Question

(5a + 4b)^2 - (3a-2b)^2

Answer 1

Answer:

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2

(5a+4b)^{2}-1(3a-2b)^{2}

(5a+4b)2−1(3a−2b)2

Simplify

1

Expand the square

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\left(5a+4b\right)^{2}-1(3a-2b)^{2}

(5a+4b)2−1(3a−2b)2

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(5a+4b)(5a+4b)-1(3a-2b)^{2}

(5a+4b)(5a+4b)−1(3a−2b)2

2

Distribute

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{\color{#c92786}{(5a+4b)(5a+4b)}}-1(3a-2b)^{2}

(5a+4b)(5a+4b)−1(3a−2b)2

5

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{\color{#c92786}{5a(5a+4b)+4b(5a+4b)}}-1(3a-2b)^{2}

5a(5a+4b)+4b(5a+4b)−1(3a−2b)2

3

Distribute

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{\color{#c92786}{5a(5a+4b)}}+4b(5a+4b)-1(3a-2b)^{2}

5a(5a+4b)+4b(5a+4b)−1(3a−2b)2

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{\color{#c92786}{25a^{2}+20ab}}+4b(5a+4b)-1(3a-2b)^{2}

25a2+20ab+4b(5a+4b)−1(3a−2b)2

4

Distribute

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25a^{2}+20ab+{\color{#c92786}{4b(5a+4b)}}-1(3a-2b)^{2}

25a2+20ab+4b(5a+4b)−1(3a−2b)2

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25a^{2}+20ab+{\color{#c92786}{20ab+16b^{2}}}-1(3a-2b)^{2}

25a2+20ab+20ab+16b2−1(3a−2b)2

5

Combine like terms

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25a^{2}+{\color{#c92786}{20ab}}+{\color{#c92786}{20ab}}+16b^{2}-1(3a-2b)^{2}

25a2+20ab+20ab+16b2−1(3a−2b)2

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25a^{2}+{\color{#c92786}{40ab}}+16b^{2}-1(3a-2b)^{2}

25a2+40ab+16b2−1(3a−2b)2

6

Expand the square

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25a^{2}+40ab+16b^{2}-1\left(3a-2b\right)^{2}

25a2+40ab+16b2−1(3a−2b)2

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Answer 2

Answer:

$=>16 a^{2}+52 a b+12 b^{2}$

Step-by-step explanation:

Given expression -

$(5a + 4b)^2 - (3a-2b)^2$

We have to simplify the above expression,

By using the binomial theorem $(p+q)^2=p^2+2pq+q^2$ to expand $(5a + 4b)^2.$

$=> 25 a^{2}+40 a b+16 b^{2}-(3 a-2 b)^{2}$

Use the binomial theorem $(p-q)^{2}=p^{2}-2 p q+q^{2}$ to expand $(3 a-2 b)^{2}.$

$=>25a^{2}+40 a b+16 b^{2}-\left(9 a^{2}-12 a b+4 b^{2}\right)$

To find the opposite of $9 a^{2}-12 a b+4 b^{2}$ find the opposite of each term.

$=>25 a^{2}+40 a b+16 b^{2}-9 a^{2}+12 a b-4 b^{2}$

Combine $25 a^{2}$ and $-9 a^{2}$ to get $16 a^{2}.$

$=>16 a^{2}+40 a b+16 b^{2}+12 a b-4 b^{2}$

Combine $40 a b$ and $12 a b$ to get $52 a b$.

$=>16 a^{2}+52 a b+16 b^{2}-4 b^{2}$

Combine $16 b^{2}$ and $-4 b^{2}$ to get $12 b^{2}$.  

$=>16 a^{2}+52 a b+12 b^{2}$

Hence, the required solution is $16 a^{2}+52 a b+12 b^{2}.$