Question

Find all the zeros for the following function y=x2 - 6x +16x-96 The zeros are​

Answer 1

The zeros for the function

`x=6,\:x=4i,\:x=-4i`

Explanation:

Given the expression

`y=x^3-6x^2+16x-96`

Plug in y = 0 to determine all the zeros

`x^3-6x^2+16x-96=0`

as

`x^3-6x^2+16x-96=\left(x-6\right)\left(x^2+16\right)`

so the expression becomes

`\left(x-6\right)\left(x^2+16\right)=0`

Using the zero factor principle

if ab=0, then a=0 or b=0 (or both a=0 and b=0)

`x-6=0\quad \mathrm{or}\quad \:x^2+16=0`

solving

`x-6=0`

`x = 6`

solving

`x^2+16=0`

subtract 16 from both sides

`x^2+16-16=0-16`

`x^2=-16`

`\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=√(f\left(a\right)),\:\:-√(f\left(a\right))`

`x=√(-16),\:x=-√(-16)`

as

`x=√(-16)`

`√(-16)=√(-1)√(16)`

as

`√(-1)=i`

so

`x=4i`

and

`x=-√(-16)`

`-√(-16)=-√(-1)√(16)`

as

`√(-1)=i`

so

`x=-4i`

Thus, the zeros for the function

`x=6,\:x=4i,\:x=-4i`