Question

Find the roots of the polynomial equation x^3-4x2+x+26=0

Answer 1

Answer:

The roots of the polynomial equation $x^{3}-4x^{2}+x+26=0$ are$x=-2, 3+2i\ and\ 3-2i$.

Step-by-step explanation:

The polynomial provided is:

$x^{3}-4x^{2}+x+26=0$

As the polynomials highest degree is 3 there will be 3 roots of the polynomial equation.

The first root can be determined by the hit-and-trial method.

For x = 2,

$x^{3}-4x^{2}+x+26=0\\(2)^{3}-4(2)^{2}+(2)+26=0\\20\neq 0$

For x = - 2

$x^{3}-4x^{2}+x+26=0\\(-2)^{3}-4(-2)^{2}+(-2)+26=0\\0=0$

Thus, one of the roots of the polynomial $x^{3}-4x^{2}+x+26=0$ is $x=-2$ or one factor is $(x+2)$.

Now divide he polynomial with this factor as follows:

$\frac{x^{3}-4x^{2}+x+26}{x+2} =x^{2}-6x+13$

The resultant polynomial is a quadratic equation.

Solve the equation $x^{2}-6x+13$ as follows:

$x=\frac{-b\pm\sqrt{b^{2}-4ac} }{2a} \\=\frac{-(-6)\pm\sqrt{(-6)^{2}-4\times1\times13} }{2\times1} \\=\frac{6\pm\sqrt{-16} }{2} \\=\frac{6\pm4i }{2} \\=3\pm2i$

Thus, the roots of the polynomial equation $x^{3}-4x^{2}+x+26=0$ are$x=-2, 3+2i\ and\ 3-2i$.