Let #f(x)=x+8# and #g(x)=2x^2-128#, how do you find #h(x)=(g(x))/(f(x))#?
1 Answer
Dec 21, 2017
Explanation:
#(g(x))/(f(x))#
#=(2x^2-128)/(x+8)#
#"factorise the numerator"#
#2(x^2-64)larrcolor(blue)"common factor of 2"#
#x^2-64" is a "color(blue)"difference of squares"#
#•color(white)(x)a^2-b^2=(a-b)(a+b)#
#"here "a=x" and "b=8#
#rArrx^2-64=(x-8)(x+8)#
#rArr(2cancel((x+8))(x-8))/cancel((x+8))#
#rArrh(x)=2(x-8)=2x-16#
