Question

The polynomial equation x^3-4x^2+2x+10=x^2-5x-3 has complex roots 3+2i. What is the other root? Use a graphing calculator and a system of equations.

Answer 1

The roots of the polynomial equation are:

-1,3+2i and 3-2i

Explanation:

We are given a polynomial equation as:

`x^3-4x^2+2x+10=x^2-5x-3`

this equation could also be written as:

`y=x^3-4x^2+2x+10----(1)`

and
`y=x^2-5x-3-----(2)`

Now if this equation has a complex root as:

`3+2i`

and we know that for any polynomial equation witth real coefficients the complex roots always appear in pair.

if a+ib is some root of a polynomial equation with real coefficients then its conjugate a-ib is also a root of the equation.

Hence, the other root of this polynomial equation will be:
`3-2i`

Now we will graph these system of equations i.e. equation (1) and equation (2).

on solving the equation we have:

`x^3-5x^2+7x+13=0\\\\(x+1)(x^2-6x+13)=0`

(since

`x^3-4x^2+2x+10=x^2-5x-3\\\\x^3-4x^2+2x+10-x^2-(-5x)-(-3)=0\\\\x^3-4x^2-x^2+2x+5x+10+3=0\\\\x^3-5x^2+7x+13=0`

)

Hence on solving the equation we have:

x=-1,3+2i and 3-2i are the roots of the equation.