# Binomial function

• Nov 21st 2006, 04:38 AM
Yppolitia
Binomial 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 AM
topsquark
Quote:

Originally Posted by Yppolitia
How would I factorize

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

and then solve

f(x)=0?

Thanks a lot.:)

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 AM
putnam120
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.