Tutorial: Solving polynomial equations
Adaptive game version
This tutorial: Part A: Solving linear and quadratic equations
(This topic is also in Section 0.5 in Finite Mathematics and Applied Calculus)
Polynomial expressions and polynomial equations
In the %%factoringquadraticstut we looked at expressions of the form
 $ax^2 + bx + c, \quad $($a \neq 0,\ b,$ and $c$ constants) Example: $\color{steelblue}{3x^24x4}$
 $3x^24x4 = (3x+2)(x2).$
Polynomial and polynomial equation
A polynomial is an algebraic expression of the form
A polynomial is an algebraic expression of the form
 $ax^n + bx^{n1} + \cdots + rx + s$
Examples
$3x2$ \gap[10] has degree 1, as the highest power of $x$ that appears with a nonzero coefficient is $x = x^1.$ Degree 1 polynomials are called linear expressions.
\\
\\ $2x  x^2$ \gap[10] has degree 2, as the highest power of $x$ that appears with a nonzero coefficient is $x^2.$ Degree 2 polynomials are called quadratics. \\
\\ $0x^4+3x^2+1$ \gap[10] also has degree 2, as the highest power of $x$ that appears with a nonzero coefficient is $x^2.$
\\
\\ $4x^3x^25$ \gap[10] has degree 3. Degree 3 polynomials are called cubics.
\\
\\ $x^41$ \gap[10] has degree 4. Degree 4 polynomials are called cuartics.
Some for you
A polynomial equation of degree $n$ is an equation that can be written in the form
 $ax^n + bx^{n1} + \cdots + rx + s = 0. \quad (a \neq 0) \quad \qquad$ Degree n polymonial = 0
Examples
$3x2 = 0$ \gap[10] is a degree 1 polymonial equation. Degree 1 polynomial equations are called linear equations.
\\
\\ $3x^22x+1 = 0$ \gap[10] is a degree 2 polymonial equation. Degree 2 polynomial equations are called quadratic equations.
\\
\\ $4x^3x^25 = 0$ \gap[10] is a degree 3 polymonial equation. Degree 3 polynomial equations are called cubic equations.
\\
\\ $x^41 = 0$ \gap[10] is a degree 4 polymonial equation. Degree 4 polynomial equations are called quartic equations.
Solving linear and quadratics equations by factoring
We have already seen how to solve linear equations, and also quadratic equations whose lefthand sides factor, in the %%factoringquadraticstut. Here we review that material:
Solution of ax + b = 0 Eg. −2x + 5 = 0
Suggested video for this topic: Video by patrickJMT
1. Subtract the $b$ from both sides: (If b is negative, this amounts to adding a number to both sides.)
$ax + b \color{red}{\ \  \ b}$ \t ${}= \color{red}{\ \  \ b}$
\\ $ax $ \t ${}=  b$
$\color{slateblue}{2x+5} \color{red}{\ \  \ 5} $ \t $\color{slateblue}{{}= } \color{red}{\ \  \ 5}$
\\ $\color{slateblue}{2x}$ \t $\color{slateblue}{{}= 5}$
2. Divide both sides by $a.$:
$\dfrac{ax}{\color{red}{a}}$ \t $= \dfrac{b}{\color{red}{a}}$
\\ $x$ \t $= \dfrac{b}{\color{red}{a}}$
$\dfrac{2x}{\color{red}{2}}$ \t ${}=\dfrac{5}{\color{red}{2}}$
\\ $x$ \t ${}=\dfrac{5}{2}$
Some for you to do
Solution of ax^{2} + bx + c = 0 when the quadratic factors Eg. 2x^{2} + 5x  3 = 0
Suggested video for this topic: Video by patrickJMT
1. Factor the lefthand side:
$\color{blue}{(px + q)}\color{red}{(rx + t)} = 0$
$\color{blue}{(2x1)}\color{red}{(x+3)} = 0$
2. As the product of the two factors is zero, one of them must be zero::
$\color{blue}{px + q = 0}$ #[or][o]# $\color{red}{rx + t= 0}$
$\color{blue}{2x1 = 0}$ #[or][o]# $\color{red}{x+3 = 0}$
3. Solve the resulting linear equation(s)::
$\color{blue}{x = \dfrac{q}{p}}$ #[or][o]# $\color{red}{x = \dfrac{t}{r}}$
$\color{blue}{x = \dfrac{1}{2}}$ #[or][o]# $\color{red}{x= 3}$
Some for you to do
Solve for $x:$ (If there is more than one solution, separate them by commas.)
Determining when a quadratic factors
%%Q:
How do I know whether I can factor a quadratic, and how do I solve a quadratic equation if it does not factor?%%A: The question has two parts. We start by answering the first part: how to recognize whether or not a quadratic equation factors.
Test for Factoring
The quadratic $ax^2 + bx + c,$ with $a, b,$ and $c$ being integers (whole numbers), factors as $(rx + s)(tx + u)$ with $r, s, t,$ and $u$ integers precisely when the quantity
 $b^2  4ac$
• If the quantity $b^24ac$ is positive but not a perfect square (for instance, $b^24ac = 15$), then the quadratic still factors as $(rx + s)(tx + u),$ but not over the integers: the numbers $s$ and $u$ will both be irrational.
• If $b^24ac$ is negative, then the quadratic does not factor at all.
• If $b^24ac$ is negative, then the quadratic does not factor at all.
Examples
1. $3x^24x+1$ has $a = 3, b = 4, c = 1,$ and so
Some for you to do
 $b^24ac = (4)^24(3)(1) = 1612 = 4,$
 $3x^24x+1 = (3x1)(x1).$
 $b^24ac = (4)^24(1)(2) = 168 = 8,$
%%A: The quadratic formula can be used to obtain any possible solutions of $ax^2+bx+c=0$ whether or not the left hand side factors over the integers:
Solving quadratic equations with the quadratic formula
Using the quadratic formula to solve quadratics (works every time)
The solutions of the quadratic equation $ax^2 + bx + c = 0$ are
 $x = \dfrac{b\pm\sqrt{b^24ac}}{2a}.$
 If $\Delta$ is positive, there are two distinct real solutions.
 If $\Delta$ is zero, there is only one real solution: $x = \dfrac{b}{2a}.$ (Why?)
 If $\Delta$ is negative, there are no real solutions.
Examples
1. $x^25x12 = 0$ has $a = 2, b = 5,$ %%and $c = 12.$ The discriminant is
$\Delta = b^24ac = (5)^24(2)(12) = 25 + 96 = 121,$
which is positive, so there are two real solutions:
$x = \dfrac{b\pm\sqrt{b^24ac}}{2a}$ \t ${}= \dfrac{5\pm\sqrt{(5)^24(2)(12)}}{2(2)}$
\\ \t ${}= \dfrac{5\pm\sqrt{121}}{4} = \dfrac{5\pm 11}{4}$
\\ \t ${}= \dfrac{16}{4}$ #[or][o]# $\dfrac{6}{4}$
\\ \t ${}= 4$ #[or][o]# $\dfrac{3}{2}$
Note that in this case the discriminant is a perfect square: $121=11^2.$ So, we could also have gotten the answer by factoring.
2. $x^2+ 2x  1 = 0$ has $a = 1, b = 2,$ %%and $c = 1.$ The discriminant is $\Delta = b^24ac = 2^24(1)(1) = 4+4 = 8,$
which is positive, so there are two real solutions:
$x = \dfrac{b\pm\sqrt{b^24ac}}{2a}$ \t ${}= \dfrac{2\pm\sqrt{2^24(1)(1)}}{2(1)}$
\\ \t ${}= \dfrac{2\pm\sqrt{8}}{2} = \dfrac{2\pm2\sqrt{2}}{2}$
\\ \t ${}= 1 + \sqrt{2}$ %%or $1  \sqrt{2}$
In this case the discriminant is not a perfect square, so the left hand side does not factor over the integers.
3. $4x^2 = 12x  9$ can be rewritten as $4x^212x+9 = 0$ which has $a = 4, b = 12,$ %%and $c = 9.$The discriminant is $\Delta = b^24ac = (12)^24(4)(9) = 144144 = 0,$
which is zero, so there is only one real solution:
$x = \dfrac{b\pm\sqrt{b^24ac}}{2a}$ \t ${}= \dfrac{12\pm\sqrt{(12)^24(4)(9)}}{2(4)}$
\\ \t ${}= \dfrac{12\pm\sqrt{144144}}{8} = \dfrac{12\pm\sqrt{0}}{8}= \dfrac{12}{8}$
\\ \t ${}= \dfrac{3}{2}$
In this case the discriminant is a perfect square: $0 = 0^2,$ so we could also have gotten the answer by factoring.
4. $x^2+x+1 = 0$ has $a = 1, b = 1,$ %%and $c = 1.$ The discriminant is $\Delta = b^24ac = 1^24(1)(1) = 1  4 = 3,$
which is negative, so there are no real solutions.
Factoring hardtofactor quadratics
The quadratic formula can also be used to factor quadratics. This is useful in cases when a quadratic that we know must factor is nonetheless difficult or tedious to factor, like, say,
$36x^2+109x+80,$
whose discriminant is$\Delta = b^2  4ac = (109)^2  4(36)(80) = 361,$
which happens to be a perfect square:
$\sqrt{361} = 19,$
whose discriminant is$\Delta = b^2  4ac = (109)^2  4(36)(80) = 361,$
meaning that it does factor, somehow or other, but the usual "trial and error" method requires looking at all combinations of factors of $36$ and $80$, and there are tons of those!
Factoring quadratics with the quadratic formula: Stef's surefire method
(Note that the video is really only for the special case in which $a, b,$ and $c$ have no common factor with $a$ positive, giving $k = 1$ in Step 3, so it is not quite right in general. I could not find a video that does it right in the general case.)
 Check that $\Delta = b^2  4ac$ is a perfect square. (If the numbers are big, use a calculator to take the square root.)
 Use the quadratic formula to get both roots in lowest terms $\dfrac{p}{q}$ and $\dfrac{r}{s}.$

The desired factorization is $k(qxp)(sxr),$ where $k = \dfrac{a}{qs}.$
($k = \pm 1$ when the original quadratic has no common integer factor. )
(Note that the video is really only for the special case in which $a, b,$ and $c$ have no common factor with $a$ positive, giving $k = 1$ in Step 3, so it is not quite right in general. I could not find a video that does it right in the general case.)
Example
Let's use this method to factor $36x^2+93x+60$.
 $a = 36, b = 93, c = 60 \ \ \Rightarrow \ \ \Delta = b^2  4ac = (93)^24(36)(60) = 9,$ which is a perfect square. ✓

Roots:
$\dfrac{b\pm\sqrt{b^24ac}}{2a} = \dfrac{93 \pm \sqrt{9}}{2(36)} = \dfrac{93 \pm 3}{72},$giving$\dfrac{90}{72} = \dfrac{5}{4} = \dfrac{p}{q} \qquad$ %%and $\qquad \dfrac{96}{72} = \dfrac{4}{3} = \dfrac{r}{s}$

$k = \dfrac{a}{qs} = \dfrac{36}{(4)(3)} = \dfrac{36}{12} = 3,$ so the desired factorization is $36x^2+93x+60$ \t ${}=k(qxp)(sxr)$ \\ \t ${}= 3(4x(5))(3x  (4))$ \\ \t ${}= 3(4x+5)(3x+4)$Done!
Now try the exercises in Section 0.5 in Finite Mathematics and Applied Calculus.
or move ahead to the next tutorial by pressing "Next tutorial" on the sidebar.
Copyright © 2022 Stefan Waner and Steven R. Costenoble