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:
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