Tutorial: Solving polynomial equations
Game version
(This topic is also in Section 0.6 in Finite Mathematics and Applied Calculus)

Equations that reduce to the form P·Q = 0
If we can manipulate a complicatedlooking 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
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
 $P \cdot Q = 0$,
 $P = 0$ or $Q = 0.$
 $P \cdot Q \cdot R = 0$,
 $P = 0$, $Q = 0$, or $R = 0$.
Examples
1. ^{ } \t $4x^7x^5 = 0$ \gap[10]
\\ \t $x^5(4x^21)=0$ \t \gap[10] Factor the lefthand side.
\\ \t Either $x^5=0$ or $(4x^21)=0.$ \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 $(2x+1)(x^24)(x3)(x^24) = 0$ \gap[10]
\\ \t $[(2x+1)  (x3)](x^24) = 0$ \t \gap[10] Common factor $(x^24)$
\\ \t $(x+4)(x^24) = 0$ \t \gap[10] Simplify.
\\ \t $(x+4)(x2)(x+2) = 0$ \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 $x\sqrt{2x1} = \sqrt{2x1}$ \gap[10]
\\ \t $x\sqrt{2x1}  \sqrt{2x1} = 0$
\\ \t $\sqrt{2x1}(x  1) = 0$ \t \gap[10] Common factor $\sqrt{2x1}$
\\ \t Either $\sqrt{2x1}=0$ or $x1=0.$ \t \gap[10] Either $P=0$ or $Q=0.$
\\ \t Either $2x1=0$ or $x1=0.$ \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 complicatedlooking 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.
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 $\displaystyle \frac{x^21}{x2} = 0$
\\ \t $x^21=0$ \t If $\dfrac{P}{Q}=0$ then $P=0$.
\\ \t $(x1)(x+1) = 0$ \t Factor.
\\ \t $x=1$ #[or][o]# $x=1$ \t
Some for you
Neither solution makes the denominator of the original expression zero, so we accept both.
\\ \\
2. ^{ } \t $\displaystyle \frac{(x+1)(x+2)^2  (x+1)^2(x+2)}{(x+1)^4  (x2)^2} = 0$
\\ \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.
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.
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