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

Printable View

- Jan 30th 2013, 04:33 AMLotrnerdfactoring 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 - Jan 30th 2013, 05:39 AMShakarriRe: 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.

Quote:

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

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. - Jan 30th 2013, 05:57 AMLotrnerdRe: 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.

- Jan 30th 2013, 06:00 AMPlatoRe: factoring question
- Jan 31st 2013, 03:38 AMShakarriRe: 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.

Its hard to write long division on these forums, take a look at this example Algebraic Long Division