Polynomials

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\framed }[0]{\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\shaded }[0]{\def \FrameCommand {\fboxsep =\FrameSep \colorbox {shadecolor}}\MakeFramed {\FrameRestore }} \newcommand {\shaded* }[0]{\def \FrameCommand {\fboxsep =\FrameSep \colorbox {shadecolor}}\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\leftbar }[0]{\def \FrameCommand {\vrule width 3pt \hspace {10pt}}\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\titled-frame }[1]{\def \FrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame {\textbf {#1}}}\def \FirstFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame [$\blacktriangleright $]{\textbf {#1}}}\def \MidFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame [$\blacktriangleright $]{\textbf {#1\ (cont)}}}\def \LastFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame {\textbf {#1\ (cont)}}}\MakeFramed {\advance \hsize -20pt \FrameRestore }} \newcommand {\FrameCommand }[0]{\setlength \fboxrule {\FrameRule }\setlength \fboxsep {\FrameSep }\fbox } \newcommand {\FirstFrameCommand }[0]{\FrameCommand } \newcommand {\MidFrameCommand }[0]{\FrameCommand } \newcommand {\LastFrameCommand }[0]{\FrameCommand } \newcommand {\FrameHeightAdjust }[0]{6pt} \newcommand {\bysame }[0]{\mbox {\rule {3em}{.4pt}}\,} \newcommand {\N }[0]{\mathbb N} \newcommand {\C }[0]{\mathbb C} \newcommand {\W }[0]{\mathbb W} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\Q }[0]{\mathbb Q} \newcommand {\R }[0]{\mathbb R} \newcommand {\A }[0]{\mathbb A} \newcommand {\D }[0]{\mathcal D} \newcommand {\F }[0]{\mathcal F} \newcommand {\ph }[0]{\varphi } \newcommand {\ep }[0]{\varepsilon } \newcommand {\aph }[0]{\alpha } \newcommand {\QM }[0]{\begin {center}{\huge \textbf {?}}\end {center}} \newcommand {\le }[0]{\leqslant } \newcommand {\ge }[0]{\geqslant } \newcommand {\a }[0]{\wedge } \newcommand {\v }[0]{\vee } \newcommand {\l }[0]{\ell } \newcommand {\mat }[0]{\mathsf } \newcommand {\vec }[0]{\mathbf } \newcommand {\subset }[0]{\subseteq } \newcommand {\supset }[0]{\supseteq } \newcommand {\emptyset }[0]{\varnothing } \newcommand {\xto }[0]{\xrightarrow } \newcommand {\qedsymbol }[0]{$\blacksquare $} \newcommand {\bibname }[0]{References and Further Reading} \newcommand {\bar }[0]{\protect \overline } \newcommand {\hat }[0]{\protect \widehat } \newcommand {\tilde }[0]{\widetilde } \newcommand {\tri }[0]{\triangle } \newcommand {\minipad }[0]{\vspace {1ex}} \newcommand {\leftexp }[2]{{\vphantom {#2}}^{#1}{#2}} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\theta }[0]{\vartheta } \newcommand {\l }[0]{\ell } \newcommand {\d }[0]{\,d} \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Problems about polynomials.

Explain what is meant by a polynomial in a variable $x$ .

Informally: A polynomial in $x$ is an algebraic expression that can be written as a sum of terms, each of which is a whole-number power of $x$ multiplied by some real number.

Formally: For a whole number $n$ , a polynomial (in $x$ ) of degree $n$ is an expression that can be written in the form

$a_nx^n+a_{n-1}x^{n-1}+\dots +a_1x + a_0$

where the $a_i$ ’s are real numbers and $a_n\ne 0$ . A polynomial function of degree $n$ is a function that can be defined by a polynomial.

Indicate the degree of the following polynomials. For expressions that are not polynomials, type NP. $\begin {array}{cc} 3x-3 & \answer {1} \\ \sqrt {x} & \answer [format=string]{NP} \\ x^{15} & \answer {15} \\ 1-x-x^2 & \answer {2} \\ 2^x + x^2 & \answer [format=string]{NP} \\ (1+x)(2+x)x & \answer {3} \\ 56 & \answer {0} \\ 2/x & \answer [format=string]{NP} \end {array}$

Given: $3x^7 -x^5 + x^4 -16x^3 + 27 = a_7 x^7 + a_6x^6 + a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x^1 + a_0$ Find $a_0$ , $a_1$ , $a_2$ , $a_3$ , $a_4$ , $a_5$ , $a_6$ , $a_7$ .

Answer: $a_0=\answer {27}$ , $a_1=\answer {0}$ , $a_2=\answer {0}$ , $a_3=\answer {-16}$ , $a_4=\answer {1}$ , $a_5=\answer {-1}$ , $a_6=\answer {0}$ , $a_7=\answer {3}$ .

Given: $6x^5+a_4 x^4 -x^2 + a_0 = a_5 x^5 - 24 x^4 + a_3 x^3 + a_2 x^2 - 5$ Find $a_0$ , $a_1$ , $a_2$ , $a_3$ , $a_4$ , $a_5$ .

Answers: $a_0=\answer {-5}$ , $a_1=\answer {0}$ , $a_2=\answer {-1}$ , $a_3=\answer {0}$ , $a_4=\answer {-24}$ , $a_5=\answer {6}$ .

Is it true that polynomials are equal if and only if their coefficients are equal? Explain your reasoning.

Yes. This is the fact we used to complete the previous two problems.

Is it true that numbers are equal if and only if their digits are equal? Explain your reasoning.

For whole numbers written in the same base, the answer is yesno .

For decimals, the answer is yesno . [Note: We will learn why and how later in the course.]

For fractions, of course, the answer is . For example, $\frac {2}{3}=\frac {4}{6}$ , and the “digits” of the fraction on the left are clearly different from those of the fraction on the right.

Explain how to add two polynomials. Explain, in particular, how “collecting like terms” is an application of the properties of arithmetic.

Use the associative and commutative properties to rearrange the sum so that like terms are consecutive. Then use the distributive property to “collect” the like terms. For example, $3x^2+4x^2 = (3+4)x^2 = 7x^2$ .

Explain how to multiply two polynomials.

Use the distributive property: Multiply each term in the first polynomial by each term in the second polynomial. Then collect like terms.

Here is an example of the polynomial division algorithm:

Describe how to perform this algorithm:

This is very much like the long division algorithm for counting numbers:

(a): Write part of the quotient;
(b): Multiply that part of the quotient by the divisor;
(c): Subtract that product from the dividend;
(d): Repeat until the remaining part of the dividend is has a degree less than the degree of the divisor.

Note that, in the last step, what counts as “less than” is different for dividing polynomials than for dividing counting numbers.

Find the quotient and divisor when dividing $x^3+4x^2-1$ by $x+2$ . Quotient: $\answer {x^2+2x-4}$ ; remainder: $\answer {7}$ .

Find the quotient and divisor when dividing $x^3+4x^2-1$ by $x^2+1$ . Quotient: $\answer {x+4}$ ; remainder: $\answer {-x-5}$ .

State the Division Theorem for polynomials. Give some relevant and revealing examples of this theorem in action.

Informally, when dividing a polynomial by a (non-zero) polynomial, we can always find a sensible quotient and remainder (both polynomials). More formally, given $n(x)$ and a non-zero divisor $d(x)$ , we can find $q(x)$ and $r(x)$ such that $n(x) = \answer {d(x)q(x)+r(x)}$ with the degree of $r(x)$ greater thanequal toless than the degree of $d(x)$ .

Write $35_{\text {ten}}$ in base two. $35_{\text {ten}} = \answer [given]{100011}_{\text {two}}$

Find a polynomial $p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1 x + a_0$ such that the $a_i$ ’s are integers greater than or equal to $0$ and less than $2$ such that $p(2) = 35$ . $p(x) = \answer [given]{1}x^5 + \answer [given]{0}x^4 + \answer [given]{0}x^3 + \answer [given]{0}x^2 + \answer [given]{1}x + \answer [given]{1}$

How are the previous two problems related?

First, we think of our polynomial $p(x)$ as an object in base $x$ . Since the coefficients are $0$ or $1$ only, when we plug in $x=2$ , we won’t have to do any rearranging between the places. So, once we plug in $x=2$ , the coefficients form a number in base $x=2$ . Plugging in $x=2$ is the process we would go through to convert from base two to base ten. So, the value of the polynomial at $x=2$ will be the base ten value of $100011_{\text {two}}$ , or $35_{\text {ten}}$ .

Consider $x^2 + 2x + 3$ . This can be thought of as a “number” in base $x$ . Express this number in base $x+1$ , that is, find $b_0$ , $b_1$ , $b_2$ such that $b_2(x+1)^2 + b_1(x+1) + b_0 = x^2 + x + 1.$ $x^2 + x + 1 = \answer [given]{1}(x+1)^2 + \answer [given]{0}(x+1) + \answer [given]{2}.$