Given #P(x) = 2x^2-x+2# and #Q(x) = 2x-1#, what is #P(Q(x))# ?

2 Answers
George C.
Nov 11, 2017

#P(Q(x)) = 8x^2-10x+5#

Explanation:

Note that changing the variable name does not really affect what the polynomial is. So we can write:

#P(t) = 2t^2-t+2#

Then, substituting #Q(x)# for #t# it may be clear that:

#P(Q(x)) = 2(Q(x))^2-Q(x)+2#

#color(white)(P(Q(x))) = 2(2x-1)^2-(2x-1)+2#

#color(white)(P(Q(x))) = 2(4x^2-4x+1)-(2x-1)+2#

#color(white)(P(Q(x))) = (8x^2-8x+2)-(2x-1)+2#

#color(white)(P(Q(x))) = 8x^2-10x+5#

Y
Nov 11, 2017

Answer: #P(Q(x))=8x^2-10x+5#

Explanation:

This is a composition of functions problem. Note that #P(Q(x))# means to substitute #Q(x)# as the "#x#" variable in the #P(x)# expression.

Therefore, given #P(x)=2x^2-x+2# and #Q(x)=2x-1#:
#P(Q(x))=P(2x-1)#
#=2(2x-1)^2-(2x-1)+2#

We can continue simplifying by noting that
#(a+-b)^2=a^2+-2ab+b^2#

So:
#2(2x-1)^2-(2x-1)+2#
#=2(4x^2-4x+1)-2x+1+2# by the distributive property
#=8x^2-8x+2-2x+3# by the distributive property again
#=8x^2-10x+5# which is our answer.

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