Tutorial: Solving polynomial equations

(This topic is also in Section 0.6 in Finite Mathematics and Applied Calculus)

Note For this tutorial, it is assumed that you know how to solve linear and quadratic equations. If you feel you need to review this, go back to the %%solvepolystut.

Equations often arise in calculus that are not polynomial equations of low degree. Many of these complicated-looking equations can be solved easily using a few basic techniques to first write them in the form $P \cdot Q = 0$ or $P/Q = 0,$ where $P$ and $Q$ are simpler expressions.

Equations that reduce to the form P·Q = 0

If we can manipulate a complicated-looking equation into the form $P \cdot Q = 0$ then we can solve it easily we you remember the following, which we used in the preceeding topic:

Solving equations that reduce to the form P·Q = 0
We use the fact that, if a product is equal to 0, then at least one of the factors must be 0. That is, if

$P \cdot Q = 0$, then either

$P = 0$ or $Q = 0.$

Note This argument applies as well to a product of three or more terms; for instance, if

$P \cdot Q \cdot R = 0$, then

$P = 0$, $Q = 0$, or $R = 0$.

Examples

1. \t $4x^7-x^5 = 0$ \gap[10] \\ \t $x^5(4x^2-1)=0$ \t \gap[10] Factor the left-hand side. \\ \t Either $x^5=0$ or $(4x^2-1)=0.$ \t \gap[10] Either $P=0$ or $Q=0.$ \\ \t $x=0, x=-\frac{1}{2}$ #[or][o]# $x=\frac{1}{2}$ \t \gap[10] Solve the individual equations. \\ \\ 2. \t $(2x+1)(x^2-4)-(x-3)(x^2-4) = 0$ \gap[10] \\ \t $[(2x+1) - (x-3)](x^2-4) = 0$ \t \gap[10] Common factor $(x^2-4)$ \\ \t $(x+4)(x^2-4) = 0$ \t \gap[10] Simplify. \\ \t $(x+4)(x-2)(x+2) = 0$ \t \gap[10] Factor the quadratic. \\ \t $x=-4, x=2$ #[or][o]# $x=-2$ \t \gap[10] $P = 0$, $Q = 0$, #[or][o]# $R = 0$ \\ \\ 3. \t $x\sqrt{2x-1} = \sqrt{2x-1}$ \gap[10] \\ \t $x\sqrt{2x-1} - \sqrt{2x-1} = 0$ \\ \t $\sqrt{2x-1}(x - 1) = 0$ \t \gap[10] Common factor $\sqrt{2x-1}$ \\ \t Either $\sqrt{2x-1}=0$ or $x-1=0.$ \t \gap[10] Either $P=0$ or $Q=0.$ \\ \t Either $2x-1=0$ or $x-1=0.$ \t \gap[10] #[If][Si]# $\sqrt{a} = 0,$ #[then][entonces]# $a = 0$ \\ \t $x=\frac{1}{2}$ #[or][o]# $x=1$ \t \gap[10] Solve the individual equations.

Some for you

Equations that reduce to the form P/Q = 0

If we can manipulate a complicated-looking equation into the form $\dfrac{P}{Q} = 0$ then we can solve it using the following observation:

Solving equations that reduce to the form P/Q = 0
We use the fact that, if $\displaystyle \frac{P}{Q}=0$, then $P=0$.

Note
The expression $\dfrac{P}{Q}$ is not defined if $Q = 0$ so you need to eliminate all solutions that make $Q$ zero.

Examples

1. \t $\displaystyle \frac{x^2-1}{x-2} = 0$ \\ \t $x^2-1=0$ \t If $\dfrac{P}{Q}=0$ then $P=0$. \\ \t $(x-1)(x+1) = 0$ \t Factor. \\ \t $x=-1$ #[or][o]# $x=1$ \t

Neither solution makes the denominator of the original expression zero, so we accept both.

\\ \\ 2. \t $\displaystyle \frac{(x+1)(x+2)^2 - (x+1)^2(x+2)}{(x+1)^4 - (x-2)^2} = 0$ \\ \t $(x+1)(x+2)^2 - (x+1)^2(x+2) = 0$ \t If $\dfrac{P}{Q}=0$ then $P=0$. \\ \t $(x+1)(x+2)[(x+2) - (x+1)] = 0$ \t Factor. \\ \t $(x+1)(x+2)(1) = 0$ \\ \t $x = -1$ #[or][o]# $x = -2$ \t

Neither solution makes the denominator of the original expression zero, so we accept both.

\\ \\ 3. \t $\displaystyle \frac{(x+1)(x+2)^2 - (x+1)^2(x+2)}{x+2} = 0$ \\ \t $(x+1)(x+2)^2 - (x+1)^2(x+2) = 0$ \t If $\dfrac{P}{Q}=0$ then $P=0$. \\ \t $x = -1$ #[or][o]# $x = -2$ \t We solved this equation in the preceding example. \\ \t $x = -1$ \t

We reject $x = -2$ as it makes the denominator of the original expression zero, so $x$ cannot be $-2$ in the original equation.

Some for you

Combining techniques

In the following you will need to use the above techniques as well as various things you learned about rational functions in the %3.

Now try the exercises in Section 0.6 in Finite Mathematics and Applied Calculus. or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.