Question 152146: show that 3x-2 is a factor of 3x^2+x^2-20x+12. thank you.
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! Note: I assume that the 3x^2 term should really be 3x^3.
Note 2: You could show that (3x-2) is a factor of (3x^3+x^2-20x+12) through long or synthetic division. Here I will show you my attempt at a proof.
I will prove this in the most elegant fashion I can.
Proof. We use a direct proof.
Predicate: (3x-2) is a factor of 3x^3+x^2-20x+12.
By definition, a cubic, if factorisable, will factor as (ax+b)(cx^2+dx+e), with {a, b, c, d, e} particular constants.
So for our predicate, in order for (3x-2) to be a factor, we need (3x-2)(cx^2+dx+e).
Moreover, (3x-2)(cx^2+dx+e)=3x^3+x^2-20x+12
and, -2e-2dx+3ex-2cx^2+3dx^2+3cx^3=3x^3+x^2-20x+12.
By definition, polynomials are equivalent if their variables' coefficients are equal.
Thus:
-2e=12 implies e=-6
-2d+3e=-20
-2c+3d=1
3c=3 implies c=1
And it follows that:
-2d-18=-20 implies d=1.
This gives a factorization of (3x-2)(x^2+x-6)=(3x-2)(x-2)(x+3). Thus we are certain that (3x-2) is a factor of polynomial (3x^3+x^2-20x+12). QED
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