Question

How do you write a polynomial in standard form given zeros -1 (multiplicity 2), -2 - i (multiplicity 1)?

Answer 1

`f(x)=x^4+6x^3+14x^2+14x+5`

Explanation:

Zeros:
`-1` multiplicity 2
`(-2-i)` multiplicity 1

If the zero is `-1`, the factor is `(x-(-1))=(x+1)`.
A multiplicity of `color(red)2` implies there are `color(red)2` factors `(x+1)` or `(x+1)(x+1)= (x+1)^color(red)2 =(x^2+2x+1)`

For the complex zero `(-2-i)`, there will also be a zero at the complex conjugate`(-2+i)`

The factors are `(x-(-2-i))` and `(x-(-2+i))`
or
`(x+2+i)(x+2-i)`
`(x^2+2x-ix+2x+4-2i+ix+2i-i^2)`
`(x^2+4x+4-(-1))`
`(x^2+4x+5)`

Multiply all factors to find the polynomial in standard form
`f(x)=(x^2+2x+1)(x^2+4x+5)=`

`x^4+4x^3+5x^2`
`color(white)(aaaa)2x^3+8x^2+10x`
`color(white)(aaaaaaaaaa)x^2+4x+5=`

`f(x)=x^4+6x^3+14x^2+14x+5`