How do you find the product ${(8 - 10 a)}^{2}$ $(8-10a)^2$ ?

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Answer 1 · 2017-06-30T10:16:56

See a solution process below:

This is a special for of a quadratic where:

${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $(a - b)^2 = a^2 - 2ab + b^2$

Substituting:

$8$ $8$ for $a$ $a$

$10 a$ $10a$ for $b$ $b$

Gives:

${(8 - b)}^{2} = 8^{2} - (2 \cdot 8 \cdot 10 a) + {(10 a)}^{2} =$ $(8 - b)^2 = 8^2 - (2 * 8 * 10a) + (10a)^2 =$

$64 - 160 a + 100 a^{2}$ $64 - 160a + 100a^2$

Or, in standard polynomial form:

$100 a^{2} - 160 a + 64$ $100a^2 - 160a + 64$

How do you find the product (8-10a)^2(8−10a)2(8-10a)^2?