Polynomial expressions and polynomial equations
In the %4 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 to do
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.
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\\ $4x^3x^25 = 0$ \gap[10] is a degree 3 polymonial equation. Degree 3 polynomial equations are called cubic equations.
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\\ $x^41 = 0$ \gap[10] is a degree 4 polymonial equation. Degree 4 polynomial equations are called quartic equations.
Solving linear and quadratic equations
We have already seen how to solve linear equations, and also quadratic equations whose lefthand sides factor, in %4. Here we review that material:
Solution of ax + b = 0 Eg. −2x + 5 = 0
\t !22! 1. Subtract the $b$ from both sides: (If b is negative, this amounts to adding a number to both sides.)
\\
\\ \t \gap[40] \t !r! $ax + b \color{red}{\ \  \ b}$ \t $= \color{red}{\ \  \ b} \qquad \qquad$ \t !r! $\color{slateblue}{2x+5} \color{red}{\ \  \ 5} $ \t $\color{slateblue}{= } \color{red}{\ \  \ 5}$
\\ \t \t !r! $ax $ \t $=  b$ \t !r! $\color{slateblue}{2x}$ \t $\color{slateblue}{= 5}$ \t \t \gap[50] \t
\\ \t
\\ \t !10! 2. Divide both sides by $a.$:
\\
\\ \t \t !r! $\frac{ax}{\color{red}{a}}$ \t $= \frac{b}{\color{red}{a}} \qquad \qquad$ \t !r! $\frac{2x}{\color{red}{2}}$ \t $=\frac{5}{\color{red}{2}}$
\\ \t \t !r! $x$ \t $= \frac{b}{\color{red}{a}} \qquad \qquad$ \t !r! $x$ \t $=\frac{5}{2}$
Suggested video for this topic: %21
Some for you to do
Solve for $x:$
Solution of ax^{2} + bx + c = 0 when the quadratic factors Eg. 2x^{2} + 5x  3 = 0
\t !22! 1. Factor the lefthand side:
\\
\\ \t \gap[40] \t $\color{blue}{(px + q)}\color{red}{(rx + t)} = 0$ \t \gap[50] \t $\color{blue}{(2x1)}\color{red}{(x+3)} = 0$
\\ \t
\\ \t !10! 2. As the product of the two factors is zero, one of them must be zero::
\\
\\ \t \gap[40] \t $\color{blue}{px + q = 0}$ %or $\color{red}{rx + t= 0}$ \t \gap[50] \t $\color{blue}{2x1 = 0}$ %or $\color{red}{x+3 = 0}$
\\ \t
\\ \t !10! 2. Solve the resulting linear equation(s)::
\\
\\ \t \gap[40] \t $\color{blue}{x = \frac{q}{p}}$ %or $\color{red}{x = \frac{t}{r}}$ \t \gap[50] \t $\color{blue}{x = \frac{1}{2}}$ %or $\color{red}{x= 3}$
Suggested video for this topic: %22
Some for you to do
Solve for $x:$ (If there is more than one solution, separate them by commas.)
%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

$3x^24x+1$ has $a = 3, b = 4, c = 1,$ and so
 $b^24ac = (4)^24(3)(1) = 1612 = 4,$
 $3x^24x+1 = (3x1)(x1).$
 $b^24ac = (4)^24(1)(2) = 168 = 8,$
%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 (works every time)
The solutions of the quadratic equation $ax^2 + bx + c = 0$ are
 $x = \frac{b\pm\sqrt{b^24ac}}{2a}.$
 If $\Delta$ is positive, there are two distinct real solutions.
 If $\Delta$ is zero, there is only one real solution: $x = \frac{b}{2a}.$ (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
 $\Delta = b^24ac = (5)^24(2)(12) = 25 + 96 = 121,$
$x = \frac{b\pm\sqrt{b^24ac}}{2a}$ \t $= \frac{5\pm\sqrt{(5)^24(2)(12)}}{2(2)}$
\\ \t $= \frac{5\pm\sqrt{121}}{4} = \frac{5\pm 11}{4}$
\\ \t $= \frac{16}{4}$ %or $\frac{6}{4}$
\\ \t $= 4$ %or $\frac{3}{2}$
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  $\Delta = b^24ac = 2^24(1)(1) = 4+4 = 8,$
$x = \frac{b\pm\sqrt{b^24ac}}{2a}$ \t $= \frac{2\pm\sqrt{2^24(1)(1)}}{2(1)}$
\\ \t $= \frac{2\pm\sqrt{8}}{2} = \frac{2\pm2\sqrt{2}}{2}$
\\ \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  $\Delta = b^24ac = (12)^24(4)(9) = 144144 = 0,$
$x = \frac{b\pm\sqrt{b^24ac}}{2a}$ \t $= \frac{12\pm\sqrt{(12)^24(4)(9)}}{2(4)}$
\\ \t $= \frac{12\pm\sqrt{144144}}{8} = \frac{12\pm\sqrt{0}}{8}= \frac{12}{8}$
\\ \t $= \frac{3}{2}$
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  $\Delta = b^24ac = 1^24(1)(1) = 1  4 = 3,$

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,

$36x^2+109x+80,$
 $\Delta = b^2  4ac = (109)^2  4(36)(80) = 361,$
 $\sqrt{361} = 19,$
 $\Delta = b^2  4ac = (109)^2  4(36)(80) = 361,$
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, 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 $\frac{p}{q}$ and $\frac{r}{s}.$

The desired factorization is $k(qxp)(sxr),$ where $k = \frac{a}{qs}.$
($k = \pm 1$ 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, 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:
 $\frac{b\pm\sqrt{b^24ac}}{2a} = \frac{93 \pm \sqrt{9}}{2(36)} = \frac{93 \pm 3}{72},$
 $\frac{90}{72} = \frac{5}{4} = \frac{p}{q} \qquad$ %and $\qquad \frac{96}{72} = \frac{4}{3} = \frac{r}{s}$

$k = \frac{a}{qs} = \frac{36}{(4)(3)} = \frac{36}{12} = 3,$ 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!
Calculator needed with ability to show fractions! (Calculators that show fractions automatically show them in lowest terms.)

Now try some of the exercises in Section 0.4 of or , or move ahead to the next tutorial by pressing on the sidebar.
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