How do you factor #5x^3-320#?

2 Answers
Mar 10, 2016

Answer:

#5x^3-320=5(x-4)(x^2+4x+16)#

Explanation:

#5x^3-320=5(x^3-64)=5(x^3-4^3)#
#x^3-y^3=(x-y)(x^2+x y+y^2)#
#5x^3-320=5(x-4)(x^2+4x+16)#

Mar 20, 2016

Answer:

#(x-4)(5)(-2+-2isqrt3)# or #(x-4)(x^2+4x+16)(5)#

Explanation:

We are given #5x^3-320# and told to factor it completely. The first thing I always go to is synthetic division. To find my divisor, I need to know the roots of this equation. The easiest way to do that is to graph it and find an #x#intercept.

graph{y=5x^3-320}

The most obvious root is #x=4#, or #x-4=0#.

Anyways, let's set this thing up!

#color(white)(4.)#|#color(white)(00)color(blue)(5)color(white)(000)color(red)(0)color(white)(000)color(green)(0)color(white)(00)color(purple)(-320)#
#color(white)(4.)#|
#4##color(white)(.)#|__#20#_#80#__#320#___
#color(white)(4.)##color(white)(00)5color(white)(0000)20color(white)(000)80color(white)(0000)0#

So, we are left with #x-4# and #5x^2+20x+80#.

Next, I want to reduce #5x^2+20x+80#, so I'm going to factor out a #5#, which leaves us with #5(x^2+4x+16)#.

Now, I wan to check if there are any roots I can use to solve this, so let's graph it and find any #x#-intercepts
graph{y=x^2+4x+16}

It takes some zooming out, but we can see that there are no #x#-intercepts! That means that are no real solution for the remainder of this problem. But, we can still factor this more (lucky us).

I'm going to use the quadratic formula. We could also complete the square, but the formula is usually easier.

#x^2+4x+16#
#a=1#
#b=4#
#c=16#

#(-b)/(2a)+-sqrt(b^2-4*a*c)/(2a)#

This becomes #(-(4))/(2(1))+-sqrt((4)^2-4*(1)*(16))/(2(1))#, which simplifies to #-2+-sqrt(-48)/2#.

If we simplify this further, we have #-2+-(sqrt(-48))/2#, which we can reduce to #-2+-2*cancel(2)*i(sqrt3)/cancel(2)#, or #-2+-2isqrt3#.

So, we have for our complete equation, #(x-4)(5)(-2+-2isqrt3)#, but the only real solution to this is #x-4#.