Senior 3 Pre-Calculus Mathematics Module 8, Lesson 3 25 Lesson 3 Remainder Theorem and Factor Theorem Objectives 1. You will learn how to divide polynomials using long division and synthetic division. 2. You will learn how to apply the remainder and factor theorems. You have had practice in adding, subtracting, and multiplying polynomials. Dividing polynomials has many important applications and is especially valuable in factoring and finding the zeros of polynonual functions There are two algorithms for polynomial division: long division and synthetic division. We will examine long division first and the relate it to synthetic division Example 1 Dividex 2 —6--xbyx ÷ 2. Solution x —3 quotient divisor x + 2)x2 — x —6 dividend V x2 +2x —3x—6 V —3x—6 00 The answer to the division is the quotient. 1. Arrange the dividend and the divisor in descending powers of the variable. Notice the powers of x go from 2— 1 —0. 2. Insert with 0 coefficients any missing terms. 3. Divide the first term of the divisor into the first term of the (2 ‘ dividend I = x I and place the answer over the second term. Lx )
20
Embed
Example 1 2. - mathmadness / FrontPagemathmadness.pbworks.com/w/file/fetch/52712137/M32s... · · 2018-04-24Remainder Theoremand Factor ... dividend = divisor x quotient + remainder
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
1. You will learn how to divide polynomials using long divisionand synthetic division.
2. You will learn how to apply the remainder and factortheorems.
You have had practice in adding, subtracting, and multiplyingpolynomials. Dividing polynomials has many importantapplications and is especially valuable in factoring and findingthe zeros of polynonual functions
There are two algorithms for polynomial division: long divisionand synthetic division. We will examine long division first andthe relate it to synthetic division
Example 1
Dividex2—6--xbyx ÷ 2.
Solution
x —3 quotient
divisor x + 2)x2— x —6 dividend
V x2+2x—3x—6
V
—3x—600
The answer to the division is the quotient.
1. Arrange the dividend and the divisor in descending powers ofthe variable. Notice the powers of x go from 2— 1 —0.
2. Insert with 0 coefficients any missing terms.
3. Divide the first term of the divisor into the first term of the(2 ‘
dividend I = x I and place the answer over the second term.Lx )
6x3+16x—19x2—4=(x—2)(6x3+7x+2)+Odividend = divisor x quotient + remainder
The above examples illustrate the Division Algorithm.It states:
If/tx) and d(x) are polynomials such that d(x) 0, and the degreeof d(x) is less than or equal to the degree offix), then there arepolynomials q(x) + r(x) such thatf(x) = d(x)q(x) +r(x) where r(x) = 0dividend = (divisor)(quotient) ÷ remainder or the degree
of r(x) is lessorthan thedegree of d(x).
f(x) =(x—a)q(x)represent the same relationship
If the remainder r(x) =0, then d(x) divides evenly into fix).When the divisor is of the form x ± a, synthetic division can beused as a shortcut for long division using polynomials.
The remainder obtained in the synthetic division process has aninteresting connection to polynomial functions.Exampie 6
4 2Evaluate fix) = x — lOx — 2x ÷4 at fi—3).Solution
f(-3) = (_34) 10(3)2 2(-3) +4= 81—90+6÷4=1
Notice fi—3) = 1 which represents the remainder fromExample 5.
If a polynomial fix)) is divided by x — a, then the remainder isr =
From the Division Algorithm, we have fix) = (x — a) q(x) + Rwhere R is the remainder.
If we fInd fia)=(a —a)g(x)+ Rfia) = R which represents the remainderWhen you are interested in the value of the remainder, you willuse the Remainder Theorem to obtain the value of theremainder.
Once you have listed the possible “a” values, you can use theRemainder Theorem and Factor Theorem and/or syntheticdivision to obtain the factors.Note: When the leading coefficient is 1, then the possible “a”values are simply the factor of the constant term.
3 2fix)=x —2x —5x+6
Because the coefficient ofx3 is 1, the possible “a” values are thefactors of 6.
a = ±1, ±2, ±3, ±6
Choose a = 1Iffla) = 0, then x — a is a factor
f(1)=13_2(1)2 —5(1)÷6=1—2—5+6=0
4
:
L
Fr
;.x—lisafactor
Use synthetic division to get the second factor:1)1 —2 —5 6
1 —1 —61-—i. —6 0
quotient isx2—x—6
Notice that this is a quadratic that will factorx2—x--6=(x--3)(x÷2)