4 Solving polynomial equations when one solution is known

In Section 3.2 we gave a formula which can be used to solve quadratic equations. Unfortunately when dealing with equations of higher degree no simple formulae exist. If one of the roots can be spotted or is known we can sometimes find the others by the method shown in the next Example.

Example 25

Let the polynomial expression $x^3-17x^2+54x-18$ be denoted by $P\left(x\right)$ . Verify that $x=4$ is a solution of the equation $P\left(x\right)=0$ . Hence find the other solutions.

Solution

We substitute $x=4$ into the polynomial expression $P\left(x\right)$ :

$P\left(4\right)=4^3-17\left(4^2\right)+54\left(4\right)-8=64-272+216-8=0$

So, when $x=4$ the left-hand side equals zero. Hence $x=4$ is indeed a solution. Knowing that $x=4$ is a root we can state that $\left(x-4\right)$ must be a factor of $P\left(x\right)$ . Therefore $P\left(x\right)$ can be re-written as a product of a linear and a quadratic term:

$P\left(x\right)=x^3-17x^2+54x-8=\left(x-4\right)\times \text{ (quadratic polynomial)}$

The quadratic polynomial has already been found in a previous task so we deduce that the given equation can be written

$P\left(x\right)=x^3-17x^2+54x-8=\left(x-4\right)\left(x^2-13x+2\right)=0$

In this form we see that    $x-4=0\text{ or }{x}^2-13x+2=0$

The first equation gives $x=4$ which we already knew.

The second equation must be solved using one of the methods for solving quadratic equations given in Section 3.2. For example, using the formula we find

So $x=12.8443$ and $x=0.1557$ are roots of $x^2-13x+2$ .

Hence the three solutions of $P\left(x\right)=0$ are $x=4$ , $x=12.8443$ and $x=0.1557$ , to 4 d.p.

Task!

Solve the equation $x^3+8x^2+16x+3=0$ given that $x=-3$ is a root.

Consider the equation $x^3+8x^2+16x+3=0$ .

Given that $x=-3$ is a root state a linear factor of the cubic:

Answer

$x^3+8x^2+16x+3=\left(x+3\right)\left(a{x}^2+bx+c\right)$

where $a,b,$ and $c$ are constants. These can be found by expanding the right-hand side.

Expand the right-hand side:

Answer

Equate coefficients of $x^3$ to find $a$ :

Answer

Equate constant terms to find $c$ :

Answer

Equate coefficients of $x^2$ to find $b$ :

Answer

This enables us to write the equation as $\left(x+3\right)\left(x^2+5x+1\right)=0$ so $x+3=0$ or $x^2+5x+1=0.$

Now solve the quadratic and state all three roots:

Answer

Exercises

1. Verify that the given value is a solution of the equation and hence find all solutions:
   1. $x^3+7x^2+11x+2=0$ , $x=-2$
   2. $2x^3+11x^2-2x-35=0$ , $x=-5$
2. Verify that $x=1$ and $x=2$ are solutions of $x^4+4x^3-17x^2+8x+4$ and hence find all solutions.

Answer