1. ## factoring question

Could someone help me with the following question?

a^3 + a^2 - a - 1

I can't seem so get to the answer. Id love to know how you arrive to the answer too.

Thanks

2. ## Re: factoring question

Using the factor theorem the roots multiplied together have to be -1 or 1 (the constant term in the equation)

This is a small proof od the factor theorem.
The cubic equation should have three roots, call these roots p, q and r
When these are the three roots
(a-p)(a-q)(a-r)= a^3 + a^2 - a - 1
a^3 - (p+q+r)*a^2 + (pq+pr+rq)*a - pqr= a^3 + a^2 - a - 1
The constant on the left hand side -pqr is equal to the constant on the right hand side -1
So pqr= 1
Whatever the values of p, q and r are they must multiply to give 1.
It is possible that they are not integers (whole numbers) but to begin with assume that they are.
The only factors of 1 are 1 and -1
Test if a=1 is a root
(1)^3 + (1)^2 - (1) -1= 0

So 1 is a root.
a=1 is a root implies (a-1) is a factor.

use long division to divide a^3 + a^2 - a - 1 by (a-1) and you will get a quadratic equation. Solve the quadratic equation to find the other two roots.

3. ## Re: factoring question

Thank you for the help. I undertand the first part, I just don't know how to do long division in this situation to get the last two roots.

4. ## Re: factoring question

Originally Posted by Lotrnerd
Could someone help me with the following question?
a^3 + a^2 - a - 1
I can't seem so get to the answer. Id love to know how you arrive to the answer too.
$a^3+a^2-a-1\\a^2(a+1)-(a+1)\\(a+1)(a^2-1)\\(a+1)^2(a-1)$

5. ## Re: factoring question

Plato's method of factorising directly works but it wont always be that straight forward. Long dividing will get the same answer but will always work.
Originally Posted by Lotrnerd
Thank you for the help. I undertand the first part, I just don't know how to do long division in this situation to get the last two roots.
Its hard to write long division on these forums, take a look at this example Algebraic Long Division