 Jean Blumer

2021-12-26

How do you find the roots of ${x}^{3}-12x+16=0$ ? Fasaniu

Explanation:
note that ${2}^{3}-12\left(2\right)+16=0$
$⇒x=2$ is a root and $\left(x-2\right)$ is a factor
$⇒{x}^{2}\left(x-2\right)+2x\left(x-2\right)-8\left(x-2\right)$
$=\left(x-2\right)\left({x}^{2}+2x-8\right)$
$=\left(x-2\right)\left(x+4\right)\left(x-2\right)$
$⇒{\left(x-2\right)}^{2}\left(x+4\right)=0$
$⇒x=2$ multiplicity of 2
and $x=-4$ multiplicity of 1. soanooooo40

Explanation:
Let $f\left(x\right)={x}^{3}-12x+16$
Then,
We see that
$f\left(2\right)={2}^{3}-12\cdot 2+16=8-24+16=0$
Therefore,
$\left(x-2\right)$ is a factor of $f\left(x\right)$
To find the other factors, we perform a long division

${x}^{3}-2{x}^{2}$
$0+2{x}^{2}-12x$
$+2{x}^{2}-4x$
$+0-8x+16$
$-8x+16$
$-8x+16$
$-0+0$
The quotient is
${x}^{2}+2x-8=\left(x-2\right)\left(x+4\right)$
Therefore,
${x}^{3}-12x+16={\left(x-2\right)}^{2}\left(x+4\right)$ nick1337

Another way....
$12x-{x}^{3}=16⇒x\left(12-{x}^{2}\right)=16$
16 factorises to 16(1)=8(2)=4(4) so we can quickly discover
whether or not the function has an integer root.
If it does, the factoring is made simple.
$x=2⇒2\left(12-4\right)=2\left(8\right)$ works
x=4 gives -16, but x=-4 gives (-4)(-4)=16
${x}^{3}-12x+16=\left(x-2\right)\left(x+4\right)\left(x+c\right)⇒-8c=16⇒c=-2$
Check that this holds.
Or

as there is no ${x}^{2}$ term.

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