# algebra

• September 15th 2009, 02:49 AM
thereddevils
algebra
Factorise $f(x)=x^4-5x^3+11x^2-7x-24$

I found $f(x)=(x+1)(x-3)(x^2-3x+6)$

Hence , solve the equation $24x^4+7x^3-11x^2+5x-1$

I can just see that the coefficient and the sign swapped .. but i am not sure what to do ?
• September 15th 2009, 08:20 AM
red_dog
Let $g(x)=24x^4+7x^3-11x^2+5x-1$.

$f\left(\frac{1}{x}\right)=\frac{1}{x^4}-\frac{5}{x^3}+\frac{11}{x^2}-\frac{7}{x}-24=$

$=-\frac{24x^4+7x^3-11x^2-5x+1}{x^4}=-\frac{g(x)}{x^4}$

Then, $g(x)=0\Rightarrow f\left(\frac{1}{x}\right)=0$.

But $f\left(\frac{1}{x}\right)=\left(\frac{1}{x}+1\righ t)\left(\frac{1}{x}-3\right)\left(\frac{1}{x^2}-\frac{3}{x}+6\right)$

Can you continue?