Tutorial: Solving polynomial equations

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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)

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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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(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 $36x^2+93x+60$.

$a = 36, b = 93, c = 60 \ \ \Rightarrow \ \ \Delta = b^2 - 4ac = (93)^2-4(36)(60) = 9,$ 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.