# How do you find the real solutions of the polynomial x^4+2x^3=12x^2+40x+32?

Mar 26, 2017

Solutions to ${x}^{4} + 2 {x}^{3} = 12 {x}^{2} + 40 x + 32$ are $x = - 2$ and $x = 4$

#### Explanation:

${x}^{4} + 2 {x}^{3} = 12 {x}^{2} + 40 x + 32$

or $f \left(x\right) = {x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32 = 0$

This will have roots as $\pm 1$, $\pm 2$, $\pm 4$, $\pm 8$, $\pm 16$ or $\pm 32$, although some of them may repeat.

Note that $x = - 2$ is a solution as $f \left(- 2\right) = 0$ and $\left(x + 2\right)$ is a factor of $f \left(x\right)$

Similarly $x = 4$ is a solution as $f \left(4\right) = 0$ and $\left(x - 4\right)$ is a factor of $f \left(x\right)$

As $\left(x + 2\right) \left(x - 4\right) = {x}^{2} - 2 x - 8$, dividing ${x}^{4} + 2 {x}^{3} - 12 {x}^{2} - 40 x - 32$ by ${x}^{2} - 2 x - 8$ we get

${x}^{2} \left({x}^{2} - 2 x - 8\right) + 4 x \left({x}^{2} - 2 x - 8\right) + 4 \left({x}^{2} - 2 x - 8\right)$ i.e.

our equation is $\left({x}^{2} + 4 x + 4\right) \left({x}^{2} + 4 x + 4\right) = 0$

or $\left(x + 2\right) \left(x - 4\right) {\left(x + 2\right)}^{2} = {\left(x + 2\right)}^{3} \left(x - 4\right) = 0$

Hence solutions to ${x}^{4} + 2 {x}^{3} = 12 {x}^{2} + 40 x + 32$ are $x = - 2$ and $x = 4$