# Thread: find linear factors of this quartic expression

1. ## find linear factors of this quartic expression

$f(x)=x^4-7x^3+9x^2+7x-10$
$f(1)=1^4-7*1^3+9*1^2+7*1-10=0$
$x-1$ is a factor, using long devision I have;
$f(x)=x^4-7x^3+9x^2+7x-10=(x-1)(x^3-6x^2+3x+10)=(x-1)g(x)$
$g(2)=2^3-^*2^2+3*2+10=0$
so $x-2$ is a factor, using long devision again I get;
$f(x)=x^4-7x^3+9x^2+7x-10=(x-1)(x-2)(x^2+4x+11+(32/x-2))$
and now I'm stuck, please help

2. You have two of them correctly. Now, since two of them can be factored out, you have a quadratic remaining which should be easier.

$\frac{x^{4}-7x^{3}+9x^{2}+7x-10}{(x-2)(x-1)}=x^{2}-4x-5$

You can factor that I am sure.

3. Thanks but I'm not sure how you have factored them out? Sorry

4. No its ok, you just divide the quartic by x-2 then divide the result buy x-1 to get that quadratic. Dont really know why but that works.