x^3 + 5x^2 - x - 5 = 0
Could someone point me in the right direction as to factor and solve for x? Thanks,
Jim
Hello,
You can see that the sum of the coefficients is 0.
Hence, 1 is a root...
-> $\displaystyle x^3+5x^2-x-5=(x-1)P(x)$
$\displaystyle \begin{aligned} x^3+5x^2-x-5 &=x^3-x^2+x^2+5x^2-x-5 \\
&=x^2(x-1)+6x^2-x-5 \\
&=x^2(x-1)+6x^2-6x+6x-x-5 \\
&=x^2(x-1)+6x(x-1)+5(x-1) \\
&=(x-1)(x^2+6x+5)
\end{aligned}$
Hence $\displaystyle x^3+5x^2-x-5=(x-1)(x+1)(x+6)$
Solve