Question

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

Answer 1

3 +or- 2i, -2

Explanation:

Looked up the answer for a test but couldn't find the right one so I just guessed and this is what the correct answer was when I went back and checked. Hope this helps!

Answer 2

The roots of the polynomial equation
`x^(3)-4x^(2)+x+26=0` are
`x=-2, 3+2i\ and\ 3-2i`.

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\\eq 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:

`(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*1*13} }{2*1} \\=(6\pm√(-16) )/(2) \\=(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`.