1. ## Algebraic Division

I hope this is in the correct section(i am in college and this is algebra..)

I know this should be simple but I really can't spot where I'm going wrong.

x^3+x^2+5x-2 is divisible by (x-2) I'm trying to completely factorise this by dividing this and then factorising the product. (this is the method i have to use for my C2 exam)

Could someone show me the step through of the division so I can see where I'm going wrong please.

Thank you

2. Originally Posted by Blu-Cherri
I hope this is in the correct section(i am in college and this is algebra..)

I know this should be simple but I really can't spot where I'm going wrong.

x^3+x^2+5x-2 is divisible by (x-2) I'm trying to completely factorise this by dividing this and then factorising the product. (this is the method i have to use for my C2 exam)

Could someone show me the step through of the division so I can see where I'm going wrong please.

Thank you
The given term isn't factorable. But if you take

$\displaystyle (x^3+x^2 -5x-2) \div (x-2)=x^2+3x+1$

3. So I did do it correctly...so how do you work out the exact solutions of the equation?

I guess i was using the wrong method completely

4. (x-2) is a factor of $\displaystyle x^3+x^2-5x-2$ as demonstrated.

If $\displaystyle f(x)=x^3+x^2-5x-2, then f(2)=0$

Therefore x = 2 is a root (solution) to the polynomial function.

Your factored equation is: $\displaystyle f(x)=(x-2)(x^2+3x+1)$

The quadratic trinomial is not factorable, so use the quadratic formula to find the other two solutions

5. Originally Posted by Blu-Cherri
So I did do it correctly...so how do you work out the exact solutions of the equation?

I guess i was using the wrong method completely
First of all: This isn't an equation but a term. A quotient. I used long division:
Code:
  (x^3 + x^2 - 5x - 2) ÷ (x - 2) = x^2 + 3x + 1
-(x^3 -2x^2)
-------------
3x^2 - 5x
-(3x^2 - 6x)
-------------
x - 2
-(x - 2)
----------
0

6. ah equation was in the question referring to f(x)=0 sorry i didn't make that clear.

Thank you both for all the information, I think i'll be posted here most of this week leading up to my C2 exam >_<