Question

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

Answer 1

Answer:

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

Step-by-step 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 $x^2-2x+26$:

$D=(-2)^2-4\cdot 1\cdot 26=4-104=-100=100i^2\\ \\\sqrt{D}=\sqrt{100i^2}=10i\\ \\x_{1,2}=\dfrac{-(-2)\pm 10i}{2\cdot 1}=\dfrac{2\pm 10i}{2}=1\pm 5i$

Hence,

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

and the initial polynomial factorization is

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