Department of Mathematics --- College of Science --- University of Utah

Mathematics 1010 online

Factoring and Solving Polynomial Equations

It's obvious that a polynomial such as

$\displaystyle p(x) = (x-1)(x-2) $

is zero when $ x=1 $ or $ x=2 $. The same fact is less obvious when we rewrite $ p $ as

$\displaystyle p(x) = x^2 - 3x + 2. $

In the first form, $ p $ is said to be factored It is easier to tell the roots of a polynomial if it is factored, but it is also possible to go the other direction: to factor a polynomial when we know one or more of its roots. Just how to go about this is described on this page.

Knowing a Real Root

Suppose we recognize, for example by trial and error, or drawing a graph, that the polynomial

$\displaystyle f(x) = x^3 - 4x^2 + 6x - 4 $

is zero when $ x=2 $. This can be easily checked:

$\displaystyle f(2) = 8 - 4\times 4 + 6 \times 2 - 4 = 0. $

The fact that $ 2 $ is a root implies that $ f $ can be written as $ f(x) = (x-2)\times g(x) $ where $ g $ is a polynomial of degree $ 2 $. As discussed elsewhere, $ g $ can be found by synthetic division :

\begin{displaymath} \begin{array}{rrrrr} 2:\quad& 1 & -4 & 6 & - 4 \\ & & 2 &-4 & 4\\ & 1 & -2 & 2& \underline{0} \\ \end{array} \end{displaymath}

This means that

$\displaystyle g(x) = x^2-2x +2. $

To find the remaining zeros of $ f $ we only have to find the zeros of $ g $. This means we have to solve the quadratic equation

$\displaystyle x^2-2x+2 = 0 $

which has the solutions

$\displaystyle x= 1 \pm i. $

Thus we see that the roots of $ f $ are $ 2 $ and $ 1\pm i $, and we were able to find all of them once we recognized one of them. In general, if we have a polynomial $ f $ and one real root we can use synthetic division to obtain a polynomial of degree one less whose roots equal the remaining roots of $ f $.

Knowing a Complex Root

Suppose the complex number $ z $ is a root of a polynomial $ p $. Then we can write $ p(x) = g(x) (x-z) $ just as above. However, the coefficients of $ g $ are complex numbers even if the coefficients of $ p $ are real numbers. The process still works, but it requires complex arithmetic, and there is a better way that uses real arithmetic only.

The key fact in this context is that conjugate complex $ \bar z $ of $ z $ is also a root of $ p $. It's a simple exercise to see that this is true by observing that the conjugate complex of a real number is the number itself, and that for any complex numbers $ u $ and $ v $

$\displaystyle \overline{uv} = \bar{u}\bar{v} \qquad\hbox{and}\qquad \overline{u+v} = \bar u + \bar v. $

Thus if $ x-z $ is a factor of $ p $ then $ x-\bar z $ is also a factor, and so is the product $ (x-z)(x-\bar z) $. This last term is a quadratic polynomial whose coefficients are real. To see this let

$\displaystyle z = a+bi \qquad\hbox{and}\qquad \bar z = a-bi. $

Then

\begin{displaymath} \begin{array}{rcl} (x- z)(x- \bar z) &=& (x-a-bi)(x-a+bi) \\... ...i)x + (a-bi)(a+bi)\\ &=& x^2 -2ax + a^2+b^2 \\ \end{array} \end{displaymath}

Hence if we know that $ a+bi $ is a root of $ p $ then $ a-bi $ is also a root and we can obtain a polynomial $ h $ satisfying

$\displaystyle p(x) = (x^2-2ax + a^2+b^2) h(x). $

The degree of $ h $ is $ 2 $ less than the degree of $ p $.

Let's look again at the above example. Essentially we work it backwards. Again, let

$\displaystyle f(x) = x^3 - 4x^2 + 6x - 4 $

We know that $ 1+i $ is a root of $ f $. Thus $ a=b=1 $ and $ x^2-2ax + a^2+b^2 = x^2-2x+2. $ So we can write

$\displaystyle f(x) = (x^2-2x+2)h(x) $

where $ h $ is a linear polynomial. It can be obtained by long division, and we know from the above discussion that $ h(x) = x-2 $

A Comprehensive Example

The following example shows all the principles described above in action. It does not address the question of how one might find a particular root. The answer to that question depends on the context in which the polynomial in question occurs. There are general purpose methods for finding a single root but they are beyond the scope of this class.

Let

$\displaystyle p(x) = 2x^5 - 9x^4 + 18x^3 - 6x^2 - 40x + 75 $

and suppose that we wish to find all roots of $ p $.

It is easily checked that $ x=-\frac{3}{2} $ is a root of $ p $. This means that $ \left(x+\frac{3}{2}\right) $ is a factor of $ p $. We find it by synthetic division :

\begin{displaymath} \begin{array}{rrrrrrr} -\frac{3}{2}:\quad& 2 & -9 & 18 & - 6... ... & 2 & -12 & 36 & -60 & 50 & \underline{0} \\ \end{array} \end{displaymath}

Hence

$\displaystyle p(x) = \left(x+\frac{3}{2} \right) \left( 2x^4-12x^3+36x^2-60x +50 \right). $

Note that the remainder of the division by $ \left(x+\frac{3}{2}\right) $ is zero (underlined in the above table) which confirms that $ x=-\frac{3}{2} $ is in fact a root. $ p $ can be rewritten (after multiplying first factor and dividing the second factor by 2) as

$\displaystyle p(x) = \left(2x+3 \right) \left( x^4-6x^3+18x^2-30x +25 \right). $

We now need to find the roots of

$\displaystyle g(x) = \left( x^4-6x^3+18x^2-30x +25 \right). $

It turns out that $ v= 1+2i $ is a root of $ g $.

Thus

$\displaystyle q(x) = (x-v)(x-\bar v) = x^2-2x+5 $

is a factor of $ g $. We find the quotient by long division :

\begin{displaymath} \begin{array}{rrrrrrrrr} &&&&&& x^2 & - 4x & + 5 \\ &&&&\l... ...le\hfill & \leaders\hrule\hfill \\ &&&&&&&0 \\ \end{array} \end{displaymath}

Again, the fact that the remainder is zero indicates that $ q $ is actually a factor of $ g $. In fact,

$\displaystyle g(x) = \left(x^2-5x+5\right)\left(x^2-4x+5\right). $

The roots of the first factor are $ 1\pm 2i $, that's how we constructed that factor. The roots of the second factor can be found by solving the quadratic equation

$\displaystyle \left(x^2-4x+5\right) = 0 $

which has the solutions $ x = 2\pm i $. The roots of our original polynomial $ p $ are, therefore:

$\displaystyle -\frac{3}{2}, \quad 1+2i, \quad 1-2i, \quad 2+i, \quad\hbox{and}\quad 2-i. $