Polynomial Equations

Solving Quadratic (degree two) Equations

Consider a quadratic function in the form

$$y = ax^2 + bx + c$$

and consider the values of $x$ for which $y=0$. This gives the quadratic equation

$$0 = ax^2 + bx + c$$

Solving this equation corresponds to finding the $x$-intercepts of the graph of the quadratic function.

For example, to solve $2x^2 +11x -4 = 2$, we first set the right-hand side of the equation equal to zero :

$$\begin{eqnarray*} 2x^2 +11x -4 & = & 2 \\ 2x^2 +11x - 6 & = & 0 \end{eqnarray*}$$

The graph of $y =2x^2 +11x - 6$ is

In solving this equation algebraically, we will be finding the $x$-intercepts of the graph--that is, where $y=0$. To do this, we could

1. factor (if possible)

2. use the quadratic formula

3. complete the square

To solve by factoring, we attempt to manufacture a product equal to 0: Can we find a product of two linear factors that, when FOILed, give $2x^2 +11x - 6$ ? We first consider all possible factors of the leading coefficient, 2, and the constant term, 6, that will multiply to give the desired first and last terms, $2x^2$ and $-6$ respectively. There are many such possibilities:

First factor	Second factor
$(2x+1)$	$(x-6)$
${\bf (2x-1)}$	${\bf (x+6)}$
$(2x+2)$	$(x-3)$
$(2x-2)$	$(x+3)$
$(2x+3)$	$(x-2)$
$(2x-3)$	$(x+2)$
$(2x+6)$	$(x-1)$
$(2x-6)$	$(x+1)$

Observe that of all 8 possibilities, only the second one gives the correct middle term, $11x$, when FOILed. Hence

$$\begin{eqnarray*} 2x^2 + 11x -6 & = & 0 \\ (2x-1)(x+6) & = & 0 \\ \mbox{~~e... ...r~~} & & x+6 = 0 \\ x = \frac12 \qquad \mbox{~~or~~} & & x= -6 \end{eqnarray*}$$

Alternatively, using the quadratic formula,

A third approach would be to complete the square.

Note that the solutions are the same regardless of the method used.

Solving Higher Degree Polynomials by Factoring

So far we have focused on solving quadratic equations (polynomial equations of degree two). We now consider the more general case of solving higher degree polynomial equations -- looking specifically at polynomial equations that can be factored. The strategy will be just like the factoring approach to solving a quadratic equation.

As in the quadratic case, solving any polynomial equation that equals 0 corresponds graphically to finding the $x$-intercepts of the graph of the polynomial function.

Consider the following example.

$$\begin{eqnarray*} x^5 - 3x & = 0 \quad & \mbox{Solving a polynomial equal to 0.... ...hat each solution corresponds to an $x$-intercept of the graph.} \end{eqnarray*}$$

Consider another example of how the factoring strategy can be used to solve a polynomial equation.

$$x^3 - 3x^2 - 4x + 12 = 0$$

First, observe that there is no factor common to all four terms to immediately factor out front. So we try factoring by grouping: group the first two terms together and factor that pair, then group the last two terms together and factor that pair. This strategy is useful only if the two resulting pairs share a common factor.

$$\begin{eqnarray*} x^3- 3x^2 - 4x + 12 & = & 0 \\ (x^3- 3x^2) - (4x - 12) & = ... ...x-3 & = & 0 \\ x = 2 \mbox{ or } x = -2 \mbox{ or } x & = & 3 \end{eqnarray*}$$