# MAT 101: Zeros of Polynomials

As we studied here, once you know how to find at least one rational zero of a polynomial using long division or synthetic division you can find the rest of the zeros of the polynomial. In this note, we study how to find a rational zero of a polynomial if there is one. Let $P(x)=a_nx^n+\cdots +a_1x+a_0$ and suppose that $P(x)$ has a rational zero $\frac{p}{q}$. This means that by factor theorem $P(x)$ has a factor $x-\frac{p}{q}$ or equivalently a factor $qx-p$. That is, $P(x)=(qx-p)Q(x)$ where $Q(x)$ is a polynomial of degree $n-1$. Let us write $Q(x)=b_{n-1}x^{n-1}+\cdots+b_1x+b_0$. Then we see that $a_n=qb_{n-1}$ and $a_0=-pb_0$. This means that $q$ is a factor of the leading coefficient $a_n$ of $P(x)$ and $p$ is a factor of the constant term $a_0$ of $P(x)$. Hence we have the Rational Zero Theorem.

Rational Zero Theorem. Let $P(x)=a_nx^n+\cdots +a_1x+a_0$ be a polynomial with integer coefficients where $a_n\ne 0$ and $a_0\ne 0$. If $P(x)$ has a rationa zero $\frac{p}{q}$ then $q$ is a factor of $a_n$ and $p$ is a factor of $a_0$.

Here is the strategy to find a rational zero of a polynomial $P(x)$.

STEP 1. Use the rational zero theorem to find the all candidates for a rational zero of $P(x)$.

STEP 2. Test each candidate from STEP 1 to see if it is a rational zero using the factor theorem. Once you find one say $\frac{p}{q}$, stop and move to STEP 3

STEP 3. Use long division or synthetic division (easier) to divide $P(x)$ by $x-\frac{p}{q}$ to find the rest of the zeros.

STEP 4. If necessary (in the event $Q(x)$ from STEP 3 has a higher degree), repeat the process $Q(x)$ from STEP 1.

Example. Find all zeros of $P(x)=2x^3+x^2-13x+6$.

Solution. $a_0=2$ has factors $\pm 1$ and $\pm 2$. $a_0=6$ has factors $\pm 1,\pm 2,\pm3\pm 6$. Thus all the candidates for a rational zero are
$$\pm 1,\pm 2,\pm 3,\pm 6,\pm\frac{1}{2},\pm\frac{2}{2}=\pm 1,\pm\frac{3}{2},\pm\frac{6}{2}=\pm 3$$
Since $P(2)=0$, 2 is a rational zero. Using long division or synthetic division we find $Q(x)=2x^2+5x-3=(2x-1)(x+3)$. Therefore, all zeros of $P(x)$ are $-3,\frac{1}{2},2$.

Example. Find all zeros of $P(x)=x^4-5x^3+23x+10$.

Solution. $a_n=1$ has factors $\pm 1$ and $a_0=10$ has factors $\pm 1, \pm 2, \pm 5, \pm10$. So all the candidates for a rational zero are
$$\pm 1, \pm 2, \pm 5, \pm10$$
Since $P(5)=0$, 5 is a rational zero. Using long division or synthetic division we find $Q(x)=x^3-5x-2$. We cannot factor this cubic polynomial readily so we repeat the process. The leading coeffient 1 has factors $\pm 1$ and the constant term $-2$ has factors $\pm 1,\pm 2$ so all the candidates for a rational zero of $Q(x)$ are $\pm 1,\pm 2$. $Q(-2)=0$ so $-2$ is a rational zero of $Q(x)$ (and hence of $P(x)$ as well). Using one’s favorite division we find the quotient $x^2-2x-1$ which has two real zeros $1\pm\sqrt{2}$. Therefore, all zeros of $P(x)$ are
$5, -2, 1\pm\sqrt{2}$.

It would be convenient if we can estimate how many positive real zeros and how many negative zeros without actually factoring the polynomial. Here is a machinary just for that.

Descartes’ Rule of Signs

Let $P(x)$ be a polynomial with real coefficients.

1. The number of positive real zeros of $P(x)$ is either equal to the number of variations in sign in $P(x)$ or is less than that by an even number.
2. The number of negative real zeros of $P(x)$ is either equal to the number of variations in sign in $P(-x)$ or is less than that by an even number.

Example. $P(x)=3x^6+4x^5+3x^3-x-3$ has one variation in sign so there is one positive real zero. $P(-x)=3x^6-4x^5-3x^3+x-3$ has three variations in sign so there can be either three negative zeros or one negative zero.

Upper and Lower Bounds for Real Zeros

Let $P(x)$ be a polynomial with real coefficients.

1. If we divide $P(x)$ by $x-a$ ($a>0$) using synthetic division and if the row that contains the quotient and remainder has no negative entry, then $a$ is an upper bound for the real zeros of $P(x)$.
2. If we divide $P(x)$ by $x-b$ ($b<0$) using synthetic division and if the row that contains the quotient and remainder has entries that are alternatively nonpositive and nonnegative, then $b$ is a lower bound for the real zeros of $P(x)$.

Example. If we divide $P(x)=3x^6+4x^5+3x^3-x-3$ by $x-1$ then
$$1|\begin{array}{cccccc} 3 & 4 & 3 & 0 & -1 & -3\\ & 3 & 7 & 10 & 10 & 9\\ \hline 3 & 7 & 10 & 10 & 9 & 6 \end{array}$$
Since the row that contains quotient and remainder has no negative entries, 1 is an upper bound for real zeros of $P(x)$. If we divide $P(x)$ by $x-(-2)$ then
$$-2|\begin{array}{cccccc} 3 & 4 & 3 & 0 & -1 & -3\\ & -6 & 4 & -14 & 28 & -54\\ \hline 3 & -2 & 7 & -14 & 27 & -57 \end{array}$$
The entries of the row that contains the quotient and remainder are alternatively nonpositive and nonnegative, so $-2$ is a lower bound for real zeros of $P(x)$. $P(x)$ in fact does not have any integer zeros but the upper and lower bounds helps us graphically locate the real zeros of $P(x)$. Also they can be used as initial estimates for Newton’s method, a method that can find approximations to real zeros of a polynomial. Figure 1 shows that there is one positive real zero and one negative real zero of $P(x)$.

Real zeros of P(x)=3x^6+4x^5+3x^3-x-3

# MAT 101: Dividing Polynomials

Polynomials are nice in the sense that they behave like numbers. For polynomials Division Algorithm works as well namely Given polynomials $P(x)$ and $D(x)\ne 0$ there exist unique polynomials $Q(x)$ and $R(x)$ such that
$$P(x)=D(x)Q(x)+R(x)$$
or
$$\frac{P(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$$
$P(x)$, $D(x)$, $Q(x)$, and $R(x)$ are called, respectively, the dividend, divisor, quptient and remainder. There are two ways to divide a polynomial by another polynomial. The first one is the familiar long division and it works the same way we do with numbers.

Example. Let $P(x)=8x^4+6x^2-3x+1$ and $D(x)=2x^2-x+2$. Find polynomials $Q(x)$ and $R(x)$ such that $P(x)=D(x)Q(x)+R(x)$.

Solution.

Long division

Hence $Q(x)=4x^2+2x$ and $R(x)=-7x+1$.

The other method is called synthetic division. This method cannot be used for any polynomial divisions, however it works great when the divisor is a linear polynomial and is easier than long division. Synthetic division uses only coefficients without including variables as shown in the following example.

Example. Using synthetic division divide $2x^3-7x^2+5$ by $x-3$.

Solution.

Synthetic division

Hence we have $Q(x)=2x^2-x-3$ and $R=-4$.

If a polynomial $P(x)$ is divided by a linear polynomial $x-c$, by division algorithm $P(x)$ can be written as
$$P(x)=(x-c)Q(x)+R$$
for some $Q(x)$ and $R$. So, $P(c)=R$ and hence we obtain the Remainder Theorem.

Remainder Theorem. If a polynomial $P(x)$ is divided by $x-c$, then the remainder is $P(c)$.

Example. Let $P(x)=3x^5+5x^4-4x^3+7x+3$. Use the remainder theorem to find the remainder when $P(x)$ is divided by $x+2$.

Solution. $R=P(-2)=5$.

As a corollary of the remainder theorem we have

Factor Theorem. $c$ is a zero of $P(x)$ if and only if $x-c$ is a factor of $P(x)$.

Example. Let $P(x)=x^3-7x+6$. Show that $P(1)=0$ and use this information to factor $P(x)$ completely.

Solution. Dividing $P(x)$ by $x-1$ (using long division or synthetic division) we find $Q(x)=x^2+x-6$ and $R=0$ (of course as we expected). So,
\begin{align*}
P(x)&=(x-1)(x^2+x-6)\\
&=(x-1)(x-2)(x+3)
\end{align*}

Example. Find a polynomial of degree 4 that has zeros $-3$, 0, 1, and 5.

Solution. Such a polynomial would have $x+3$, $x$, $x-1$, and $x-5$ for its factors by the factor theorem. So the simplest one is
$$P(x)=(x+3)x(x-1)(x-5)=x^4-3x^3-13x^2+15x$$

# MAT 101: One-to-One Functions and Inverse Functions

A function $y=f(x)$ is said to be one-to-one if satisfies the property
$$f(x_1)=f(x_2) \Longrightarrow x_1=x_2$$
or equivalently
$$x_1\ne x_2 \Longrightarrow f(x_1)\ne f(x_2)$$
for all $x_1,x_2$ in the domain. In plain English what this says is no two numbers in the domain are corresponded to the same number in the range. Figure 1 is the graph of $f(x)=x^2$. It is not one-to-one.

Figure 1. The graph of y=x^2

For example, $-1\ne 1$ but $f(-1)=1=f(1)$.

Figure 2. The graph of y=x^3

Figure 2 is the graph of $f(x)=x^3$. It is one-to-one as seen clearly from the graph. But let us pretend that we don’t know the graph but want to prove that it is one-to-one following the definition. Here we go. Suppose that $f(x_1)=f(x_2)$. Then $x_1^3=x_2^3$ or $x_1^3-x_2^3=(x_1-x_2)(x_1^2+x_1x_2+x_2^2)=0$. This means $x_1=x_2$ which completes the proof.

Why do we care about one-to-one functions? The reason is that if $y=f(x)$ is one-to-one, it has an inverse function $y=f^{-1}(x)$.
\begin{align*}
x&\stackrel{f}{\longrightarrow} y\\
x&\stackrel{f^{-1}}{\longleftarrow} y
\end{align*}
Given a one-to-one function $y=f(x)$, here is how to find its inverse function $y=f^{-1}(x)$

STEP 1. Swap $x$ and $y$ in $y=f(x)$. The reason we are doing this is that $\mathrm{Dom}(f)=\mathrm{Range}(f^{-1})$ and $\mathrm{Dom}(f^{-1})=\mathrm{Range}(f)$.

STEP 2. Solve the resulting expression $x=f(y)$ for $y$. That is the inverse function $y=f^{-1}(x)$.

Example. Find the inverse function of $f(x)=\frac{2x+3}{x-1}$. (It is a one-to-one function.)

Solution. STEP 1. Let $y=\frac{2x+3}{x-1}$ and swap $x$ and $y$. Then we have
$$x=\frac{2y+3}{y-1}$$

STEP 2. Let us solve $x=\frac{2y+3}{y-1}$ for $y$. First multiply $x=\frac{2y+3}{y-1}$ by $y-1$. Then we have $x(y-1)=2y+3$ or $xy-x=2y+3$. Isolating the terms that contain $y$ in the LHS, we get $xy-2y=x+3$ or $(x-2)y=x+3$. Finally we find $y=\frac{x+3}{x-2}$. This is the inverse function.

$y=f(x)$ and its inverse $y=f^{-1}(x)$ satisfy the following properties.
$$(f\circ f^{-1})(x)=x,\ (f^{-1}\circ f)(x)=x$$
The reason for these properties to hold is clear from the definition of an inverse function. We can check the properties using the above example. I will do $(f\circ f^{-1})(x)=x$ and leave the other for an exercise.
\begin{align*}
(f\circ f^{-1})(x)&=f(f^{-1}(x))\\
&=f\left(\frac{x+3}{x-2}\right)\\
&=\frac{2\left(\frac{x+3}{x-2}\right)+3}{\left(\frac{x+3}{x-2}\right)-1}\\
&=x
\end{align*}
The graph of $y=f(x)$ and the graph of its inverse $y=f^{-1}(x)$ satisfy a nice symmetry, namely they are symmetric about the line $y=x$. This symmetry helps us obtain the graph of $y=f^{-1}(x)$ when the explicit expression for $f^{-1}(x)$ is not available. You will see such a case later when you study the logarithmic functions. Figure 3 shows the symmetry with $y=x^2$ ($x\geq 0$) and its inverse $y=\sqrt(x)$.

Figure 3. The symmetry between the graphs of y=x^2 (red) and y=sqrt(x) (blue) about y=x (green)

# MAT 101: Combining Functions

It’s quite interesting that functions can be treated like numbers, namely you can define $+$, $-$, $\times$, and $\div$ on a collection of functions. How do we do this? For instance given two functions $f$ and $g$, we can define a new function $f+g$ by
$$(f+g)(x)=f(x)+g(x)$$
for all $x$ in the domain (for sake of simplicity we assume that $f$ and $g$ both have the same domain. If not, one can take the intersection of the domains of $f$ and $g$, no big deal). In a similar manner, we can also define $f-g$, $fg$, and $\frac{f}{g}$ respectively as
\begin{align*}
(f-g)(x)&=f(x)-g(x)\\
(fg)(x)&=f(x)g(x)\\
\left(\frac{f}{g}\right)(x)&=\frac{f(x)}{g(x)}\ \mbox{provided}\ g(x)\ne 0
\end{align*}

Example. Let $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$.

(a) Find the functions $f+g$, $f-g$, $fg$ and $\frac{f}{g}$ and their domains.

(b) Find $(f+g)(4)$, $(f-g)(4)$, $(fg)(4)$, $\left(\frac{f}{g}\right)(4)$

Solution. (a) $\mathrm{Dom}(f)=\{x|x\ne 2\}$ and $\mathrm{Dom}(g)=\{x|x\geq 0\}$. So the intersection is $\{x|0\leq x<2\}\cup\{x|x>2\}=[0,2)\cup(2,\infty)$ and this is the domain of $f+g$, $f-g$ and $fg$. For $\frac{f}{g}$ since $g$ is not defined at $x=0$, its domain should be $(0,2)\cup(2,\infty)$.
\begin{align*}
(f+g)(x)&=\frac{1}{x-2}+\sqrt{x}\\
(f-g)(x)&=\frac{1}{x-2}-\sqrt{x}\\
(fg)(x)&=\frac{\sqrt{x}}{x-2}\\
\left(\frac{f}{g}\right)(x)&=\frac{1}{(x-2)\sqrt{x}}
\end{align*}

(b) I will do only $(f+g)(4)$. One way to evaluate $(f+g)(4)$ is to use $(f+g)(x)$ we obtained in part (a) i.e. $(f+g)(4)=\frac{1}{4-2}+\sqrt{4}=\frac{5}{2}$. Another way is evaluating $f(4)$ and $g(4)$ first which are $f(4)=\frac{1}{2}$ and $g(4)=2$. Then $(f+g)(4)=f(4)+g(4)=\frac{5}{2}$.

Composite Functions

Given two functions $f$ and $g$, if the range of $f$ is a subset of the domain of $g$, then we can combine the two functions to create a new function which we will denote by $g\circ f$.
$$x\stackrel{f}{\longmapsto} f(x)\stackrel{g}{\longmapsto} g(f(x))$$
The above diagram hints us that we can define a new function $g\circ f$ by
$$(g\circ f)(x)=g(f(x))$$
We call $g\circ f$ “$f$ followed by $g$.”

Example. Let $f(x)=x^2$ and $g(x)=x-3$.

(a) Find composite functions $f\circ g$ and $g\circ f$ and their domains.

(b) Find $(f\circ g)(5)$ and $(g\circ f)(5)$.

Solution. (a) By definition $(f\circ g)(x)=f(g(x))=f(x-3)=(x-3)^2$. Also by definition $(g\circ f)(x)=g(f(x))=g(x^2)=x^2-3$. From these we can clearly see both their domains are $(-\infty,\infty)$. In general $\mathrm{Dom}(f\circ g)=\mathrm{Dom}(g)$ and $\mathrm{Dom}(g\circ f)=\mathrm{Dom}(f)$.

(b) $(f\circ g)(5)$ can be evaluated using $(f\circ g)(x)$ we obtained in part (a).
$$(f\circ g)(5)=(5-3)^2=4$$
There is another way to do this. If you don’t have to find $(f\circ g)(x)$ but only need to calculate $(f\circ g)(5)$, this may be simpler. First note $(f\circ g)(5)=f(g(5))$. $g(5)=5-3=2$, so $f(g(5))=f(2)=2^2=4$. Similarly we find $(g\circ f)(5)=22$. In general $(f\circ g)(x)\ne (g\circ f)(x)$.

# MAT 101: Nonlinear Inequalities

Nonlinear inequalities may seem more complicated and difficult to solve than linear inequalities. However it is not really the case. There is one simple way to solve a nonlinear inequality. It’s called the test point method. I will explain this with an example.

Example. Solve the inequality $x^2\leq 5x-6$.

Solution. The inequality can be rewritten $x^2-5x+6\leq 0$. First we find points at which $x^2-5x+6=0$. Since $x^2-5x+6=(x-2)(x-3)$, $x=2,3$. These two points divide the real line into 3 regions, where $x<2$, where $2<x<3$, and where $x>3$ as shown in Figure 1.

Figure 1. Quadratic Inequality

In each region we pick a test point to see if that test number satisfies the given inequality. If it does, any other number in the same region would satisfy the inequality. If not, any other number in the same region wouldn’t either. While this is pretty cool, you may wonder why this works. One number speaks for the entire numbers in the same region. It’s hard to explain here though but it is due to the continuity of the function $f(x)=x^2-5x+6$. I will leave it at that and will not delve into that any further. You will understand what I said when you learn calculus. In the region $x<2$, I would pick $x=0$ for a test point. But $x=0$ won’t satisfy the inequality as the LHS is $6>0$. Move onto the next region $2<x<3$. I pick $x=2.5=\frac{5}{2}$. Since $\left(\frac{5}{2}\right)^2-5\frac{5}{2}+6=\frac{25-50+24}{4}=-\frac{1}{4}<0$. So this means that $2<x<3$ is a solution of the inequality. In the final region $x>3$ I pick $x=4$. $(4)^2-5(4)+6=2>0$ so no number in this region would satisfy the inequality. Since $x=2$ and $x=3$ also satisfy the inequality, the overall solution is $2\leq x\leq 3$ or $[2,3]$ in interval notation.

The inequalities like one we just did is called quadratic inequalities. For quadratic inequalities we can actually classify solutions depending on inequalities without going through the test point method every time. Let us assume that the quadratic function $f(x)=ax^2+bx+c$ with $a>0$ has two real solutions $\alpha$ and $\beta$ ($\alpha<\beta$). Then the graph of $f(x)$ would look like one in Figure 2. You can find the solution of each of the following quadratic inequalities easily from the graph in Figure 2.

Figure 2. Quadratic Inequality

1. $ax^2+bx+c>0$: $x<\alpha$ or $x>\beta$. In interval notation, $(-\infty,\alpha)\cup(\beta,\infty)$.
2. $ax^2+bx+c\geq 0$: $x\leq\alpha$ or $x\geq\beta$. In interval notation, $(-\infty,\alpha]\cup[\beta,\infty)$.
3. $ax^2+bx+c<0$: $\alpha<x<\beta$. In interval notation, $(\alpha,\beta)$.
4. $ax^2+bx+c\leq 0$: $\alpha\leq x\leq\beta$. In interval notation, $[\alpha,\beta]$.

Let us go over a couple more examples of nonlinear inequalities that are not quadratic inequalities.

Example. Solve $x(x-1)^2(x-3)<0$.

Solution. The method is the same as the first example. We use the test point method. First find $x$ at which $x(x-1)^2(x-3)=0$. They are $x=0, 1, 3$. So there are 4 regions under consideration. $x<0$, $0<x<1$, $1<x<3$, and $x>3$. In the region where $x<0$, the test point $x=-1$ results the sign of the LHS is $+$. So $x<0$ is not a solution. In the region $0<x<1$, the test point $x=\frac{1}{2}$ results the sign of the LHS $-$, so $0<x<1$ is a solution. In the region $1<x<3$, the test point $x=2$ results the sign of the LHS still $-$. This is actually due to $(x-1)^2$. In general if you see an even number of repeated term $x-a$ in your polynomial inequality like the one we have the sign of the polynomial does not change at $x=a$. A little goody to know so you can save time. Let us move onto next and last one. For $x>3$ the test point $x=4$ results the sign of the LHS $+$, so $x>3$ is not a solution. Therefore, the overall solution is $0<x<1$ or $1<x<3$. In interval notation, it is $(0,1)\cup(1,3)$.

Example. Solve $\frac{1+x}{1-x}\geq 1$.

Solution. First rewrite the inequality as $\frac{1+x}{1-x}-1\geq 0$ which simplifies to $\frac{2x}{1-x}\geq 0$. Inequality like this we consider points at which the numerator is 0 and also points at which the denominator is 0. In our case they are $x=0, 1$ and these two points divide the real line into three regions: $x<0$, $0<x<1$, $x>1$. In the region $x<0$, the test point $x=-1$ results the sign of the LHS $-$, so $x<0$ is not a solution. In the region $0<x<1$, the test point $x=\frac{1}{2}$ results the sign of the LHS $+$, so $0<x<1$ is a solution. Finally in the region $x>1$ the test point $x=2$ results the sign of the LHS $-$, so $x>1$ is not a solution. Since $x=0$ also satisfies the inequality, the overall solution is $0\leq x<1$ or $[0,1)$ in interval notation.