How do you factor #4x^4 - 64 #?

1 Answer
Sep 3, 2016

Answer:

#4(x-2)(x+2)(x^2+4)#

Explanation:

The first step is to take out a #color(blue)"common factor"# of 4

#rArr4(x^4-16)........ (A)#

Now #x^4-16 " is a " color(blue)"difference of squares"# and in general factorises as.

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))........ (B)#

#x^4=(x^2)^2" and " 16=(4)^2rArra=x^2" and " b=4#

substitute these values into (B)

#rArrx^4-16=(x^2-4)(x^2+4)#

Note that #x^2-4# is also a #color(blue)"difference of squares"#

and using the method of above

#rArrx^2-4=(x-2)(x+2)#

substituting the values from the 2 results into (A) remembering the factor #(x^2+4)#

#rArr4x^4-64=4(x-2)(x+2)(x^2+4)#