Tutorial: Solving polynomial equations

This tutorial: Part A: Solving linear and quadratic equations

Go to Part B: Solving cubic and higher order polynomial equations

(This topic is also in Section 0.5 in Finite Mathematics and Applied Calculus)

Note For this tutorial, it is assumed that you know how to factor quadratic expressions. If you feel you need to review this, go back to the %%factoringquadraticstut.

Polynomial expressions and polynomial equations

In the %%factoringquadraticstut we looked at expressions of the form

Example:

which are called quadratic expressions or just quadratics. We also factored many of them as products of the form $(dx + e)(fx + g)$: For instance,

$3x^2-4x-4 = (3x+2)(x-2).$The factors $3x+2$ and $x-2$ are examples of linear expressions. In general, linear and quadratic expressions are examples of polynomials:

Polynomial and polynomial equation
A polynomial is an algebraic expression of the form

$ax^n + bx^{n-1} + \cdots + rx + s$where $a, b, \dots r$ and $s$ are constants, called the coefficients of the polynomial. The largest exponent of $x$ appearing in the expression with a nonzero coefficient is called the degree of the polynomial.

Examples

$3x-2$ \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 $x^2.$ \\ \\ $4x^3-x^2-5$ \gap[10] has degree 3. Degree 3 polynomials are called cubics. \\ \\ $x^4-1$ \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

Degree n polymonial = 0

Examples

$3x-2 = 0$ \gap[10] is a degree 1 polymonial equation. Degree 1 polynomial equations are called linear equations. \\ \\ $3x^2-2x+1 = 0$ \gap[10] is a degree 2 polymonial equation. Degree 2 polynomial equations are called quadratic equations. \\ \\ $4x^3-x^2-5 = 0$ \gap[10] is a degree 3 polymonial equation. Degree 3 polynomial equations are called cubic equations. \\ \\ $x^4-1 = 0$ \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 left-hand sides factor, in the %%factoringquadraticstut. Here we review that material:

Solution of ax + b = 0 Eg. −2x + 5 = 0

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 ${}=\dfrac{-5}{2}$

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Some for you to do

Solution of ax² + bx + c = 0 when the quadratic factors Eg. 2x² + 5x - 3 = 0

1. Factor the left-hand side:

$\color{blue}{(px + q)}\color{red}{(rx + t)} = 0$

$\color{blue}{(2x-1)}\color{red}{(x+3)} = 0$

2. As the product of the two factors is zero, one of them must be zero::

$\color{blue}{px + q = 0}$ #[or][o]# $\color{red}{rx + t= 0}$

$\color{blue}{2x-1 = 0}$ #[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]# $\color{red}{x= -3}$

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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

$b^2 - 4ac$ is a perfect square (that is, it is the square of an integer). When this happens, we say that the quadratic factors over the integers.

• If the quantity $b^2-4ac$ is positive but not a perfect square (for instance, $b^2-4ac = 15$), 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 $b^2-4ac$ is negative, then the quadratic does not factor at all.

Examples

1. $3x^2-4x+1$ has $a = 3, b = -4, c = 1,$ and so

$b^2-4ac = (-4)^2-4(3)(1) = 16-12 = 4,$ which is a perfect square: $4 = 2^2.$ Therefore, the quadratic does factor over the integers. In fact,

$3x^2-4x+1 = (3x-1)(x-1).$

2. $x^2-4x+2$ has $a = 1, b = -4, c = 2,$ and so

$b^2-4ac = (-4)^2-4(1)(2) = 16-8 = 8,$ which is not a perfect square: $8$ cannot be written as the squre of a whole number. Therefore, this cuadratic does not factor over the integers,

Some for you to do

%%Q: OK that answers the first part of my question above: knowing when a quadratic $ax^2+bx+c$ factors over the integers, and how to solve $ax^2+bx+c=0$ when it does. What about the second part of the question?: How do I solve $ax^2+bx+c=0$ if the left hand side does not factor over the integers?
%%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

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$ We call the quantity $\Delta = b^2-4ac$ (which we have already seen before!) the discriminant of the quadratic ($\Delta$ is the Greek letter delta) and we have the following general principle:

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.

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Examples

1. $x^2-5x-12 = 0$ has $a = 2, b = -5,$ %%and $c = -12.$ The discriminant is

$\Delta = b^2-4ac = (-5)^2-4(2)(-12) = 25 + 96 = 121,$

which is positive, so there are two real solutions:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ \t ${}= \dfrac{5\pm\sqrt{(-5)^2-4(2)(-12)}}{2(2)}$ \\ \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

$\Delta = b^2-4ac = 2^2-4(1)(-1) = 4+4 = 8,$

which is positive, so there are two real solutions:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ \t ${}= \dfrac{-2\pm\sqrt{2^2-4(1)(-1)}}{2(1)}$ \\ \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 $4x^2-12x+9 = 0$ which has $a = 4, b = -12,$ %%and $c = 9.$The discriminant is

$\Delta = b^2-4ac = (-12)^2-4(4)(9) = 144-144 = 0,$

which is zero, so there is only one real solution:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ \t ${}= \dfrac{12\pm\sqrt{(-12)^2-4(4)(9)}}{2(4)}$ \\ \t ${}= \dfrac{12\pm\sqrt{144-144}}{8} = \dfrac{12\pm\sqrt{0}}{8}= \dfrac{12}{8}$ \\ \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

$\Delta = b^2-4ac = 1^2-4(1)(1) = 1 - 4 = -3,$

which is negative, so there are no real solutions.

%%Note The solutions of the equation $ax^2 + bx + c = 0$ are also known as the roots of the quadratic $ax^2 + bx + c.$

Factoring hard-to-factor 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 sure-fire method

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 $k(qx-p)(sx-r),$ where $k = \dfrac{a}{qs}.$
($k = \pm 1$ when the original quadratic has no common integer factor. )

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Example

Let's use this method to factor $36x^2+93x+60$.

$a = 36, b = 93, c = 60 \ \ \Rightarrow \ \ \Delta = b^2 - 4ac = (93)^2-4(36)(60) = 9,$ which is a perfect square. ✓
Roots:
$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a} = \dfrac{93 \pm \sqrt{9}}{2(36)} = \dfrac{-93 \pm 3}{72},$
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 ${}=k(qx-p)(sx-r)$ \\ \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.