Let f(x)=x+8f(x)=x+8 and g(x)=2x^2-128g(x)=2x2−128, how do you find h(x)=(g(x))/(f(x))h(x)=g(x)f(x)?
1 Answer
Dec 21, 2017
Explanation:
(g(x))/(f(x))g(x)f(x)
=(2x^2-128)/(x+8)=2x2−128x+8
"factorise the numerator"factorise the numerator
2(x^2-64)larrcolor(blue)"common factor of 2"2(x2−64)←common factor of 2
x^2-64" is a "color(blue)"difference of squares"x2−64 is a difference of squares
•color(white)(x)a^2-b^2=(a-b)(a+b)∙xa2−b2=(a−b)(a+b)
"here "a=x" and "b=8here a=x and b=8
rArrx^2-64=(x-8)(x+8)⇒x2−64=(x−8)(x+8)
rArr(2cancel((x+8))(x-8))/cancel((x+8))
rArrh(x)=2(x-8)=2x-16
