Question

How do you find the roots of `x^3-12x+16=0`?

Answer 1

`x=2" with multiplicity 2"`
`x=-4" with multiplicity 1"`

Explanation:

`"note that " 2^3-12(2)+16=0`

`rArrx=2" is a root and " (x-2)" is a factor"`

`rArrx^2(x-2)+2x(x-2)-8(x-2)`

`=(x-2)(x^2+2x-8)`

`=(x-2)(x+4)(x-2)`

`rArr(x-2)^2(x+4)=0`

`rArrx=2" multiplicity of 2"`

`"and " x=-4" multiplicity of 1"`

Answer 2

The roots are `4` and `2`(double root)

Explanation:

Let `f(x)=x^3-12x+16`

Then,

We see that

`f(2)=2^3-12*2+16=8-24+16=0`

Therefore,

`(x-2)` is a factor of `f(x)`

To find the other factors, we perform a long division

`color(white)(aaaa)` `x-2` `color(white)(aaaa)` `|` `x^3+0x^2-12x+16` `color(white)(aaaa)` `|` `x^2+2x-8`

`color(white)(aaaaaaaaaaaaaa)` `x^3-2x^2`

`color(white)(aaaaaaaaaaaaaaa)` `0+2x^2-12x`

`color(white)(aaaaaaaaaaaaaaaaa)` `+2x^2-4x`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `+0-8x+16`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaa)` `-8x+16`

`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaa)` `-0+0`

The quotient is

`x^2+2x-8=(x-2)(x+4)`

Therefore,

`x^3-12x+16=(x-2)^2(x+4)` graph{x^3-12x+16 [-16.15, 15.88, -3.97, 12.05]}