How do you FOIL $(7x^7-9)(x^8-9)$ ?

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Answer 1 · 2015-06-19T18:22:07

$(7x^7-9)(x^8-9) = F + O + I + L$

$=(7x^7*x^8)+(7x^7*-9)+(-9*x^8)+(-9*-9)$

$=7x^15-63x^7-9x^8+81$

$=7x^15-9x^8-63x^7+81$

FOIL helps you to enumerate all the terms to add when multiplying two binomials.

First: $7x^7*x^8 = 7x^15$
Outside: $7x^7*-9 = -63x^7$
Inside: $-9*x^8 = -9x^8$
Last: $-9*-9 = 81$

So:

$(7x^7-9)(x^8-9) = F + O + I + L$

$=7x^15-63x^7-9x^8+81$

$=7x^15-9x^8-63x^7+81$

...using $x^a*x^b = x^(a+b)$

Answer 2 · 2015-06-19T18:22:53

I found:
$=7x^15-9x^8-63x^7+81$

Have a look:
enter image source here

How do you FOIL (7x^7-9)(x^8-9)(7x^7-9)(x^8-9)?