Post on 25-May-2015
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THE REMAINDER THEOREM
THE REMAINDER THEOREM
INTRODUCTION
In this section, The Remainder Theorem provides us with a very interesting test to determine whether a polynomial in a form x-c divides a polynomial f(x) or simply not.
SECTION OBJECTIVES, YOU SHOULD BE ABLE TO:
Understand the proving of The Remainder Theorem.
Determine if the polynomial f(x) is divisible or not divisible by polynomial of the form x-c.
LESSON PROPERTHE REMAINDER THEOREM
The remainder obtained in dividing f(x) by x-c is the value of the polynomial f(x) for x=c, that is f(c).
PROOF:
Since the divisor is of the first degree, the remainder will be a constant r. Let q(x) be the quotient, we have the identity.
f(x) = (x-c)q(x)+r.
On substituting the number c in place of x into this identity we must get equal numbers.
Since r is a constant, it is not affected by this substitution and the value of the right-hand member for x=c will be
(c-c)q(c)+r = r.
Where the value of the left hand member is f(c); hence, r=f(c) which means also that identically in x,
f (x) = (x-c)q(x) + f (c).
It follows from this theorem that f (x) is divisible by x-c if and only if f(c) = 0. QED
f(x) = A0Xn + A1Xn-1 + . . . + An
Where:
A is constantX is variablen is the number of exponent
First we let f(x) = A0Xn + A1Xn-1 + . . . + An
Then we multiply 1/ x-c , Where x-c is a linear polynomial which means the number of
exponent in x is one or the highest degree of a term is one and c is a constant.
If you don’t understand well the first one, here’s another one to understand the proving of The Remainder Theorem.
(1/ x-c)[f(x) = A0Xn + A1Xn-1 + . . . + An](1/x-c)
Then,
f(x) = A0Xn + A1Xn-1 + . . . + An
x-c x-c
Recall this, b/a ; b = aq + r
b -> dividenda -> divisorq -> quotientr -> remainder
in divisibility form.
Then
f(x)/x-c ; f(x) = (x-c)q(x) + r
f(x) -> dividendx-c -> divisorq(x) -> quotientr -> remainder
Then the form goes like this; f(x) = (x-c)q(x) + r then we substitute c in x.
f(c) = (c-c) q(c) + r
f(c) = r then we substitute f(c) in r in the original equation. f(x) = (x-c) q(x) + f(c) Then:
If f(c) = 0, then it is divisibleIf f(c) 0, then it is not divisible. QED
Remainder Theorem
If a polynomial f(x) is divided by x-r, the remainder is equal to the value of the polynomial where r is substituted for x. Divide the polynomial by x-r until the remainder, which may be zero is independent of x. Denote the quotient by Q(x) and the remainder by R. Then according to the meaning of the division,
f(x) = (x-r) Q(x) + R.
If you still don’t understand it, we have the last one different way of proving The Remainder Theorem. And we’re hoping that you will finally understand the Remainder Theorem, and ready to proceed in the examples below.
Since this an identity in x, it is satisfied by all values of x, and if we set x=r we find that,
f(r) = (r-r) Q(r) + R = 0 . Q(r) + R = R.
Here it is assumed that a polynomial is finite for every finite value of the variable. Consequently, since Q(x) is a polynomial, Q(r) is a finite number, and 0. Q(r) = 0.
EXAMPLES
Note: The solutions of the following examples are given step by step in order that any student (fast or slow learners) could be able to understand the process.
EXAMPLE 1Let f(x) = Xn+An, where n is a
positiveinteger, using Remainder Theorem, show that (x+a) | f(x) whenever n is odd.
Note:x + a = 0 ; x = -af(x) = (x+a); then substitute –a in
x.
SOLUTION:By The Remainder Theorem we know that (x+a) |f(x) if
and only iff(-a) = (-a)n + an = 0
But if n is odd positive number then there is a positive
number msuch that n= 2m + 1.
Hence,
f(-a) = (-a)2m+1 + an
= -a (-a)2m + an
= -a2m+1 + an
= -an + an
= 0
EXAMPLE 2Show that f(x) = x3 + x2 – 5x + 3 is
divisible by x + 3.SOLUTION:
So, f(-3) = (-3)3 + (-3)2 – 5 (-3) + 3 = 0 = -27 + 9 + 15 + 3 = -27 +27 = 0
Then it is divisible by x+3.
EXAMPLE 3
Under what conditions is xn+cn divisible by x+c?
SOLUTION:f(x) = Xn+cn
xn+cn x3+c3 = (x+c) (x2-cx+c2) x+c f(-c) = (-c)n + cn
n xn + cn
1 x + c2 x2 + c2
3 x3 + c3
4 x4 + c4
For n; odd- Divisible by x + c.
For n; even- Remainder is 2cn
EXAMPLE 4Using Remainder Theorem show that
x4+3x3+3x2+3x+2 is divisible by x+2. SOLUTION: f(-2) = (-2)4+3(-2)3+3(-2)2+3(-2)+2
= 16-24+12-6+2= 16+12+2-30= 30-30= 0. Then it is divisible by x+2.
EXAMPLE 5Without actual division, show that x5-
3x4+x2-2x-3 is divisible by x-3.
SOLUTION: f(3) = (3)5-3(3)4+(3)2-2(3)-3
= 243-243+9-6-3=0. Then it is divisible by x-3.
Are the lessons clear?If yes then you may
proceed to the practice exercises.
If you think you cannot still manage, please go over the
examples once more.
PRACTICE EXERCISES1. Show x3–6x2+11x-6 that is divisible by x-1 2. Use the Remainder Theorem to determine whether x = –
4 is a solution of x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8 3. x4+7x3+3x2-63x-108 is divisible by x+3. 4. X3+4x2-9x-36 is divisible by x-3. 5. X3+4x2-9x-36 is divisible by x-1.
I can do it alone.
Check your answers next
page “BE HONEST”
ANSWERS:1. Show that x3–6x2+11x-6 is divisible by x-1 Solution:
X=1
= (1)3-6(1)2+11(1)-6= 1-6+11-6= -5+5= 0 .
2. x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8; x= -4
Solution:
= (-4)6 + 5(-4)5 + 5(-4)4 + 5(-4)3 + 2(-4)2 – 10(-4) – 8= 4096-5120+1280-320+32+40-8=-1024+960+72-8= -64+64= 0.
3. x4+7x3+3x2-63x-108; x=-3
Solution:
= (-3)4+7(-3)3+3(-3)2-63(-3)-108= 81-189+27+189-108= -108+216-108= 108-108= 0.
4. X3+4x2-9x-36; x=3
Solution:= (3)3+4(3)2-9(3)-36= 27+36-27-36= 63-63= 0.
5. X3+4x2-9x-36; x=1
Solution:
= (1)3+4(1)2-9(1)-36= 1+4-9-36= 5-45= 40. X-1 does not divide X3+4x2-9x-36.
ASSIGNMENT
1. Find the remainder when 4x3 – 5x + 1 is divided by:
a) x – 2 b) x + 3c) 2x – 1
2. The expression 4x2 – px + 7 leaves a remainder of –2 when divided by x – 3. Find the value of p.
REFERENCES
http://www.purplemath.com/modules/remaindr.htm
http://www.onlinemathlearning.com/remainder-theorem.html
http://www.wyzant.com/help/math/algebra/remainder_theorem
Marvin Marcus, Henryk Minc. College Algebra. USA. Houghton Mifflin Company, 1970.
Rider, Paul R. College Algebra.