I need help with this one: x^3 +x^2 - 14x -24

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- Feb 21st 2013, 07:32 PMdelilahFactoring a quadrinomial?
I need help with this one: x^3 +x^2 - 14x -24

- Feb 21st 2013, 07:40 PMjakncokeRe: Factoring a quadrinomial?
by inspection i note that $\displaystyle (-2)^3 + (-2)^2 - 14 * (-2) - 24 = -8 + 4 + 28 - 24 = 0 $

so (x+2) is a factor of this polynomial. Dividing $\displaystyle x^3 + x^2 - 14 - 24 $ by (x+2) i get $\displaystyle (x+2)(x^2-x-12) $ using quadratic formula for the 2nd deg, i get roots x = 4, x = -3, so the full factorization is (x+2)(x-4)(x+3) = $\displaystyle x^3 + x^2 - 14x - 24 $ - Feb 21st 2013, 07:47 PMdelilahRe: Factoring a quadrinomial?
Thanks for the quick reply. I haven't done this for years, and am in the process of re-certifying as a teacher (but factoring quadrinomials is not something I have to do for my job in general). Can you explain how one thinks about roots when working with problems like this one? DId you list the factors of 24 and then start testing those in the problem, replacing for x? Or is there a different way to think about how you solved it?

- Feb 21st 2013, 08:12 PMjakncokeRe: Factoring a quadrinomial?
There is a theorem that says that if a is a root for a polynomial, (x-a) is a factor of the polynomial

so lets take the degree 3 polynomial you have here. $\displaystyle p(x) = a_3x^3 + a_2x^2 + a_1x + a_0 $ that is the general form a degree 3 polynomial, where $\displaystyle a_3,a_2,..$ represent some number. so if a value b was a root for p(x), then p(b) = 0. By the theorem (x-b)*(some other polynomial) = p(x).

you can find the some other polynomial by dividing p(x) by x-b.