How would I factorize

x^3+2x^2-9x-18

and then solve

f(x)=0?

Thanks a lot.:)

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- Nov 21st 2006, 04:38 AMYppolitiaBinomial function
How would I factorize

x^3+2x^2-9x-18

and then solve

f(x)=0?

Thanks a lot.:) - Nov 21st 2006, 05:02 AMtopsquark
Factor by grouping: (Typically if you are asked to factor a cubic it will be a variation of this method.)

$\displaystyle (x^3 + 2x^2) - (9x + 18)$

$\displaystyle x^2(x + 2) - 9(x + 2)$

$\displaystyle (x^2 - 9)(x + 2)$

Now note that $\displaystyle x^2 - 9$ factors:

So we get:

$\displaystyle (x + 3)(x - 3)(x + 2)$

So if we have $\displaystyle f(x) = x^3+2x^2-9x-18 = 0$

Then

$\displaystyle (x + 3)(x - 3)(x + 2) = 0$

So x = -3, -2, or 3.

-Dan - Nov 21st 2006, 05:44 AMputnam120
also if you do not know how to factor that well you could try and find one root, with the rational root theorem.

lets say that u have $\displaystyle f(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$ now since this has a root lets call it $\displaystyle r$ we then have

$\displaystyle f(x)=(x-r)Q(x)$

where $\displaystyle Q(x)=y_nx^{n-1}+y{n-1}x^{n-1}+\dots+y_2x+y_1$. as you notice the degree of Q is one less than that of f.