How do you multiply #(8x + 3) ( 8x - 3)#?

2 Answers
Jade
Nov 21, 2016

#64x^2-9#

Explanation:

F.O.I.L

First, Outer, Inner, Last

Multiply the First terms in each set of parenthesis

#8x xx8x=64x^2#

Multiply the Outermost terms in each set of parenthesis

#8x xx -3=-24x#

Multiply the Inner most terms in each set of parenthesis

#3 xx 8x=24x#

Multiply the Last terms in each set or parenthesis

#3xx-3=-9#

#64x^2-24x+24x-9#

Combine like terms

#64x^2-9#

This is DTS, or Difference of Two Squares

Because the first term, #64#, can be square rooted and the last term, #9#, can be square rooted

Meave60
Nov 21, 2016

#(8x+3)(8x-3)=64x^2-9#

Explanation:

#(8x+3)(8x-3)#

This is an example of a difference of squares: #(a^2-b^2)=(a+b)(a-b)#, where #a=8x# and #b=3#.

#(8x)^2-(3^2)#

#64x^2-9#

You could also use the FOIL method, though it is more work. However, you should know both procedures in case your instructor indicates a specific method.

http://www.mesacc.edu/~scotz47781/mat120/notes/polynomials/foil_method/foil_method.html

#(8x+3)(8x-3)=8x*8x+8x*-3+3*8x+3*-3#

#(8x+3)(8x-3)=64x^2-24x+24x-9#

#(8x+3)(8x-3)=64x^2-cancel(24x)+cancel(24x)-9#

#(8x+3)(8x-3)=64x^2-9#

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