## Tutorial: Solving polynomial equations

* Adaptive game version*

(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

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:
*P*·*Q*= 0**Solving equations that reduce to the form**

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

- $P \cdot Q = 0$,

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

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

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

- $P = 0$, $Q = 0$, 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

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

*P*/*Q*= 0We 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.*August 2022*

Copyright © 2022 Stefan Waner and Steven R. Costenoble