Question

For the polynomial below, -1 is a zero of multiplicity of two.

h(x) = x^4+23x^2+50x+26 . express h(x) as a product of linear factors.

Answer 1

`h(x)=(x+1)^2(x-1-5i)(x-1+5i)`

Explanation:

Given that -1 is a zero of multiplicity of two for the polynomial
`h(x) = x^4+23x^2+50x+26.` This means that
`x-(-1)=x+1` is twice the linear factor.

Rewrite the polynomial
`h(x) = x^4+23x^2+50x+26` as follows:

`h(x)\\ \\= x^4+23x^2+50x+26\\ \\=x^4+x^3-x^3-x^2+24x^2+24x+26x+26\\ \\=x^3(x+1)-x^2(x+1)+24x(x+1)+26(x+1)\\ \\=(x+1)(x^3-x^2+24x+26)\\ \\=(x+1)(x^3+x^2-2x^2-2x+26x+26)\\ \\=(x+1)(x^2(x+1)-2x(x+1)+26(x+1))\\ \\=(x+1)(x+1)(x^2-2x+26)`

Find linear factors of the quadratic polynomial
`x^2-2x+26`:

`D=(-2)^2-4\cdot 1\cdot 26=4-104=-100=100i^2\\ \\√(D)=√(100i^2)=10i\\ \\x_(1,2)=(-(-2)\pm 10i)/(2\cdot 1)=(2\pm 10i)/(2)=1\pm 5i`

Hence,

`x^2-2x+26=(x-(1+5i))(x-(1-5i))=(x-1-5i)(x-1+5i)`

and the initial polynomial factorization is

`h(x)=(x+1)^2(x-1-5i)(x-1+5i)`