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

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Answer 1 · 2017-03-26T10:19:06

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

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

or $f(x)=x^4+2x^3-12x^2-40x-32=0$

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

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

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

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

$x^2(x^2-2x-8)+4x(x^2-2x-8)+4(x^2-2x-8)$ i.e.

our equation is $(x^2+4x+4)(x^2+4x+4)=0$

or $(x+2)(x-4)(x+2)^2=(x+2)^3(x-4)=0$

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

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