If #p(x) = x^2 -1# and #q(x) = 5(x-1)#, what is #(p-q)(x)#?

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Answer 1 · 2017-12-13T00:30:06

#x^2-5x+4#

#(p-q) (x)=p(x)-q(x)=(x^2-1)-5*(x-1)=x^2-5x+4#

Answer 2 · 2017-12-13T00:34:30

#x^2- 5x+4#

Question, #(p-q)(x)#

Here both #p# and #q# are functions assuming #(p-q)(x)# means #p(x)-q(x)#.

#(p-q)(x) = #
#p(x)-q(x)#

#= p(x) - q(x)#
#= (x^2-1) - {5(x-1)}#
#= x^2-1 - {5x-5}#
#= x^2-1 - 5x+5#
#= x^2- 5x+4#

Thus, #(p-q)(x)=x^2- 5x+4#.