How do you factor: #y= 32x^3 - 4 #?

2 Answers
Apr 14, 2017

Answer:

#y=4(2x-1)(4x^2+2x+1)#

Explanation:

Recall:

#(a^3-b^3)=(a-b)(a^2+ab+b^2)#

Here we go.

#y=32x^3-4#

By factoring out #4#,

#Rightarrow y=4(8x^3-1)#

By rewriting a bit,

#Rightarrow y=4((2x)^3-1^3)#

By applying the formula at the top with #a=2x# and #b=1#,

#Rightarrow y=4(2x-1)((2x)^2+2x cdot 1+1^2)#

By cleaning up a bit,

#y=4(2x-1)(4x^2+2x+1)#

I hope that this was clear.

Apr 14, 2017

Answer:

#y=4(2x-1)(4x^2+2x+1)#

Explanation:

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

#rArry=4(8x^3-1)to(1)#

now #8x^3-1# is a #color(blue)"difference of cubes"# and factorises in general as.

#color(red)(bar(ul(|color(white)(2/2)color(black)(a^3-b^3=(a-b)(a^2+ab+b^2))color(white)(2/2)|)))#

#8x^3=(2x)^3" and " 1^3=1#

#"using " a=2x" and " b=1#

#rArr8x^3-1=(2x-1)(4x^2+2x+1)#

#"going back to " (1)#

#rArry=4(2x-1)(4x^2+2x+1)#