How do you find the product #(y-2)(y+4)#?

1 Answer
Feb 15, 2017

Answer:

#color(blue)(y² + 2y -8)#

Explanation:

using the Distributive Property of Real Numbers,
FOIL METHOD (First Term, Outer Term, Inner Term, and the Last Term)

#color(blue)((a+b)(a+b) = a² + 2ab + b²)#
#color(blue)((a-b)(a+b) = a² - b²)#

Answering using the Distributive Property #color(red)((FOIL))#
= #(y−2)(y+4)#
First Term,
= #(color(blue)(y## − 2)##(color(blue)(y)+4)#
Outer Term,
= #(color(blue)(y## − 2)(y+##color(blue)(4)#)
Inner Term,
= #(y##color(blue)(−2)##)(##color(blue)(y)##+4)#
Last Term,
=#(y##color(blue)(-2)#)#(y##color(blue)(+4))#

First Term, #y# and #y#
= #y * y# = #color(blue)(y²)#

Outer Term, #y# and #4#
= #y * 4# = #color(blue)(4y)#

Inner Term, #-2# and #y#
= #-2 * y# = #color(blue)( -2y)#

Last Term, #-2# and #+4#
= #-2 * 4# = #color(blue)(-8)#

Plugging all the answers, we get:

= #color(blue)(y² + 4y - 2y -8)#

always simplify the answer by combining (adding) like terms, the like term is the #4y# and #-2y#, we get #4y + (-2y)# = #2y#

so the final answer is,

= #color(blue)(y² + 2y -8)#