Practice questions
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Chapter 4: Polynomial and rational functions
Try each problem first. Then check your work with the solution.
1. Polynomial functions and their graphs.
(a) Describe the end behavior of ${P(x)=-x^4+3x^3-2}$.
(b) Find the zeros of ${P(x)=x(x-2)^2(x+1)^3}$ and their multiplicities. At which zeros does the graph cross the ${x}$-axis?
(c) Sketch the graph of ${P(x)=-x^3+x^2+6x}$.
(d) Show that ${P(x)=x^3+2x-5}$ has a zero between ${1}$ and ${2}$.
Solution:
(a)
The leading term is ${-x^4}$. The degree is even, and the leading coefficient is negative. So both ends go down:
${P(x)\to-\infty}$ as ${x\to\pm\infty}$.
(b)
The zero ${0}$ has multiplicity ${1}$, the zero ${2}$ has multiplicity ${2}$, and the zero ${-1}$ has multiplicity ${3}$.
The graph crosses the axis at zeros of odd multiplicity: ${0}$ and ${-1}$. At ${2}$, the multiplicity is even, so the graph only touches the axis and turns back.
(c)
Factor out ${-x}$, then factor the quadratic:
$\begin{align*}&P(x)\\&=-x(x^2-x-6)\\&=-x(x-3)(x+2)\end{align*}$
The zeros are ${-2}$, ${0}$, and ${3}$, each of multiplicity ${1}$. The leading term is ${-x^3}$, so the left end goes up and the right end goes down. Test values:
| Test ${x}$ | ${P(x)}$ | Graph |
|---|---|---|
| ${-3}$ | ${18}$ | above |
| ${-1}$ | ${-4}$ | below |
| ${1}$ | ${6}$ | above |
| ${4}$ | ${-24}$ | below |
(d)
${P(1)=1+2-5=-2}$, and ${P(2)=8+4-5=7}$. The signs are opposite. By the intermediate value theorem, ${P}$ has a zero between ${1}$ and ${2}$.
2. Dividing polynomials.
(a) Divide ${3x^3-2x^2+4x-3}$ by ${x^2+1}$.
(b) Use synthetic division to divide ${x^4-3x^3+5x-6}$ by ${x-2}$.
(c) Use synthetic division to find ${P(-1)}$ for ${P(x)=2x^4-x^3+3x-1}$.
(d) Show that ${x+3}$ is a factor of ${P(x)=x^3+4x^2+x-6}$. Then factor ${P(x)}$ completely.
Solution:
(a)
Use long division. Write the missing terms with ${0}$:
| ${3x}$ | ${-2}$ | |||
| ${x^2+1}$ | ${3x^3}$ | ${-2x^2}$ | ${+4x}$ | ${-3}$ |
| ${-}$ | ${3x^3}$ | ${+0x^2}$ | ${+3x}$ | |
| ${-2x^2}$ | ${+x}$ | ${-3}$ | ||
| ${-}$ | ${-2x^2}$ | ${+0x}$ | ${-2}$ | |
| ${x}$ | ${-1}$ |
The quotient is ${3x-2}$, and the remainder is ${x-1}$:
$\begin{align*}&3x^3-2x^2+4x-3\\&=(x^2+1)(3x-2)+x-1\end{align*}$
(b)
Here ${c=2}$. There is no ${x^2}$ term, so the coefficients are ${1}$, ${-3}$, ${0}$, ${5}$, ${-6}$:
| ${2}$ | ${1}$ | ${−3}$ | ${0}$ | ${5}$ | ${−6}$ |
| ${2}$ | ${−2}$ | ${−4}$ | ${2}$ | ||
| ${1}$ | ${−1}$ | ${−2}$ | ${1}$ | ${−4}$ |
The quotient is ${x^3-x^2-2x+1}$, and the remainder is ${-4}$.
(c)
Divide by ${x-(-1)}$, so ${c=-1}$. The coefficients are ${2}$, ${-1}$, ${0}$, ${3}$, ${-1}$:
| ${−1}$ | ${2}$ | ${−1}$ | ${0}$ | ${3}$ | ${−1}$ |
| ${−2}$ | ${3}$ | ${−3}$ | ${0}$ | ||
| ${2}$ | ${−3}$ | ${3}$ | ${0}$ | ${−1}$ |
By the remainder theorem, ${P(-1)=-1}$.
Check: ${P(-1)=2+1-3-1=-1}$.
(d)
${P(-3)=-27+36-3-6=0}$. So ${x+3}$ is a factor, by the factor theorem. Divide by ${x+3}$, with ${c=-3}$:
| ${−3}$ | ${1}$ | ${4}$ | ${1}$ | ${−6}$ |
| ${−3}$ | ${−3}$ | ${6}$ | ||
| ${1}$ | ${1}$ | ${−2}$ | ${0}$ |
The quotient is ${x^2+x-2=(x+2)(x-1)}$. So
${P(x)=(x+3)(x+2)(x-1)}$.
3. Zeros of polynomials.
(a) List the possible rational zeros of ${P(x)=3x^3-x^2-8x-4}$.
(b) Find all zeros of ${P(x)=3x^3-x^2-8x-4}$.
(c) Find all complex zeros of ${P(x)=x^4-x^3-x^2-x-2}$.
(d) Find a polynomial of degree ${4}$ with real coefficients and leading coefficient ${1}$. It has the zeros ${3i}$ and ${1}$, where ${1}$ has multiplicity ${2}$.
Solution:
(a)
The factors of the constant term ${4}$ are ${\pm1}$, ${\pm2}$, ${\pm4}$. The factors of the leading coefficient ${3}$ are ${\pm1}$, ${\pm3}$. So the candidates ${\dfrac{p}{q}}$ are
$\pm1,\ \pm2,\ \pm4,\ \pm\dfrac{1}{3},\ \pm\dfrac{2}{3},\ \pm\dfrac{4}{3}$.
(b)
${P(1)=3-1-8-4=-10}$, so ${1}$ is not a zero. Try ${2}$:
| ${2}$ | ${3}$ | ${−1}$ | ${−8}$ | ${−4}$ |
| ${6}$ | ${10}$ | ${4}$ | ||
| ${3}$ | ${5}$ | ${2}$ | ${0}$ |
The remainder is ${0}$, so ${2}$ is a zero. Factor the quotient:
${3x^2+5x+2=(3x+2)(x+1)}$
The zeros are ${2}$, ${-\dfrac{2}{3}}$, and ${-1}$.
(c)
The possible rational zeros are ${\pm1}$ and ${\pm2}$. ${P(2)=16-8-4-2-2=0}$, so ${2}$ is a zero:
| ${2}$ | ${1}$ | ${−1}$ | ${−1}$ | ${−1}$ | ${−2}$ |
| ${2}$ | ${2}$ | ${2}$ | ${2}$ | ||
| ${1}$ | ${1}$ | ${1}$ | ${1}$ | ${0}$ |
The quotient is ${x^3+x^2+x+1}$. Factor it by grouping:
$\begin{align*}&x^2(x+1)+(x+1)\\&=(x+1)(x^2+1)\end{align*}$
So ${P(x)=(x-2)(x+1)(x^2+1)}$. The zeros of ${x^2+1}$ are ${\pm i}$. All zeros: ${2}$, ${-1}$, ${i}$, ${-i}$.
(d)
The coefficients are real, so the conjugate ${-3i}$ is also a zero. With ${1}$ counted twice, that makes ${4}$ zeros. The pair ${\pm 3i}$ gives ${(x-3i)(x+3i)=x^2+9}$. So
$\begin{align*}&P(x)\\&=(x^2+9)(x-1)^2\\&=(x^2+9)(x^2-2x+1)\\&=x^4-2x^3+x^2+9x^2-18x+9\\&=x^4-2x^3+10x^2-18x+9\end{align*}$
4. Rational functions.
(a) Find the asymptotes and the intercepts of ${r(x)=\dfrac{2x-6}{x+1}}$.
(b) Find the holes and the asymptotes of ${r(x)=\dfrac{x^2-4}{x^2+x-2}}$.
(c) Find the asymptotes of ${r(x)=\dfrac{2x^2+3x-1}{x+2}}$.
(d) Sketch the graph of ${r(x)=\dfrac{x}{x^2-4}}$.
Solution:
(a)
The denominator is ${0}$ at ${x=-1}$, so ${x=-1}$ is a vertical asymptote. Both degrees are ${1}$, so the horizontal asymptote is ${y=\dfrac{2}{1}=2}$.
The numerator is ${0}$ at ${x=3}$: the ${x}$-intercept is ${3}$. The ${y}$-intercept is ${r(0)=\dfrac{-6}{1}=-6}$.
(b)
Factor:
${r(x)=\dfrac{(x-2)(x+2)}{(x+2)(x-1)}}$
The factor ${x+2}$ cancels, so there is a hole at ${x=-2}$. In lowest terms, ${r(x)=\dfrac{x-2}{x-1}}$. At ${x=-2}$, this gives ${\dfrac{-4}{-3}=\dfrac{4}{3}}$. So the hole is at ${\left(-2,\dfrac{4}{3}\right)}$.
The vertical asymptote is ${x=1}$. Both degrees are ${2}$, so the horizontal asymptote is ${y=1}$.
(c)
The vertical asymptote is ${x=-2}$. The numerator has degree one more than the denominator, so divide. Use synthetic division with ${c=-2}$:
| ${−2}$ | ${2}$ | ${3}$ | ${−1}$ |
| ${−4}$ | ${2}$ | ||
| ${2}$ | ${−1}$ | ${1}$ |
So ${r(x)=2x-1+\dfrac{1}{x+2}}$. The slant asymptote is ${y=2x-1}$.
(d)
The denominator factors as ${(x-2)(x+2)}$. So the vertical asymptotes are ${x=\pm 2}$. The numerator has a lower degree, so the horizontal asymptote is ${y=0}$.
The only intercept is ${(0,0)}$. The function is odd, since ${r(-x)=-r(x)}$. Test values:
| Test ${x}$ | ${r(x)}$ | Graph |
|---|---|---|
| ${-3}$ | ${-0.6}$ | below |
| ${-1}$ | ${\approx 0.33}$ | above |
| ${1}$ | ${\approx -0.33}$ | below |
| ${3}$ | ${0.6}$ | above |