How do you multiply $(7n^2+8n+7)(7n^2+n-5)$ ?

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Answer 1 · 2017-02-28T16:36:17

$49n^4 +63n^3 + 22n^2 = 33n - 35$

Distribution-
$(7n^2 + 8n + 7)(7n^2 + n - 5)$

Multiply $7n^2$ , $8n$ , and $7$ EACH by $7n^2$ , $n$ , and $-5$ .

For $7n^2$ :
$7n^2*7n^2 = 49n^4$
$7n^2*n = 7n^3$
$7n^2*-5 = -35n^2$

do the same for $8n$ and $7$
You should get:
$(49n^4 + 7n^3 - 35n^2) + (56n^3 +8n^2 - 40n) +(49n^2 +7n - 35)$

Now combine like terms-

$49n^4 + 63n^3 + 22n^2 - 33n - 35$

How do you multiply (7n^2+8n+7)(7n^2+n-5)(7n^2+8n+7)(7n^2+n-5)?