1. ## College Algebra Question

I have the answers to this question but as I am looking through my notes I am confused at how my professor got the answers. If someone could explain it to me would really help me out -- Thankyou

the Problem is x3-36x=16(x-6)

So far I have :

x(x2-36)=16(x-6)
x(x-6)(x+6)=16(x-6)

and I get confused on what to do next ....

2. Originally Posted by Yup318
I have the answers to this question but as I am looking through my notes I am confused at how my professor got the answers. If someone could explain it to me would really help me out -- Thankyou

the Problem is x3-36x=16(x-6)

So far I have :

x(x2-36)=16(x-6)
x(x-6)(x+6)=16(x-6)

and I get confused on what to do next ....
Put everything onto one side and factor out (x-6):

$\displaystyle (x-6)([x(x+6)]-16) = 0$

We know that either (x-6) = 0 or x(x+6)-16=0.
Therefore one solution is x=6

Expand the other factor to give $\displaystyle x^2+6x-16=0$

Then solve using your favourite method.

I get the roots to be -8, 2 and 6