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Copyright © by Houghton Mifflin Company, Inc. All rights reserved Example: Divide & Check Example: Divide x 2 + 3x – 2 by x – 1 and check the answer. x x 2 + x 2x2x– 2 2x + 2 – 4– 4 remainder Check: correct (x + 2) quotient (x + 1) divisor + (– 4) remainder = x 2 + 3x – 2 dividend Answer: x – 4– 4
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Division of Polynomials
Digital Lesson
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Dividing Polynomials
Long division of polynomials is similar to long division of whole numbers.
dividend = (quotient • divisor) + remainder
The result is written in the form:
quotient +divisor
remainder divisor dividend
When you divide two polynomials you can check the answer using the following:
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+ 2 2 3 1 2 xxx
Example: Divide x2 + 3x – 2 by x – 1 and check the answer.
x
x2 + x2x – 22x + 2
– 4
remainder
Check:
xxxxx
22 1.
xxxx 2)1(2.
xxxxx 2)()3( 22 3.
22 2 xxxx4.
22)1(2 xx5.
4)22()22( xx6.
correct(x + 2)quotient
(x + 1)divisor
+ (– 4)remainder
= x2 + 3x – 2 dividend
Answer: x + 2 +1x
– 4
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Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer.
1 4 0 2 2 2 23 xxxx Write the terms of the dividend in descending order.
23
22 x
xx
1.
x2
232 22)22( xxxx 2.
2x3 – 2x2
2233 2)22(2 xxxx 3.
2x2 + 4x
xxx
22 2
4.
+ x
xxxx 22)22( 2 5.
2x2 – 2x
xxxxx 6)22()42( 22 6.
6x – 1
326
xx
7.
+ 3
66)22(3 xx8.
6x – 6
remainder5)66()16( xx9.
5
Check: (x2 + x + 3)(2x – 2) + 5 = 4x + 2x3 – 1
Answer: x2 + x + 3 22
x5
Since there is no x2 term in the dividend, add 0x2 as a placeholder.
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6 5 2 2 xxxx
x2 – 2x– 3x + 6
– 3
– 3x + 60
Answer: x – 3 with no remainder.
Check: (x – 2)(x – 3) = x2 – 5x + 6
Example: Divide x2 – 5x + 6 by x – 2.
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Example: Divide x3 + 3x2 – 2x + 2 by x + 3 and check the answer.
2 2 3 3 23 xxxx
x2
x3 + 3x2
0x2 – 2x
– 2
– 2x – 68
Check: (x + 3)(x2 – 2) + 8 = x3 + 3x2 – 2x + 2
Answer: x2 – 2 + 3x 8
+ 2
Note: the first subtraction eliminated two terms from the dividend.
Therefore, the quotient skips a term.
+ 0x
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16
Synthetic division is a shorter method of dividing polynomials. This method can be used only when the divisor is of the form x – a. It uses the coefficients of each term in the dividend.
Example: Divide 3x2 + 2x – 1 by x – 2 using synthetic division.
3 2 – 12
Since the divisor is x – 2, a = 2.
3
1. Bring down 32. (2 • 3) = 6
6
8 15
3. (2 + 6) = 84. (2 • 8) = 165. (–1 + 16) = 15
coefficients of quotient remainder
value of a coefficients of the dividend
3x + 8Answer: 2
x15
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Example: Divide x3 – 3x + 4 by x + 3 using synthetic division.
Notice that the degree of the first term of the quotient is one less than the degree of the first term of the dividend.
remainder
)3()43( 3 xxx
a
coefficients of quotient
– 3
Since, x – a = x + 3, a = – 3.
1 0 – 3 4
1
– 3
– 3
9 – 18
6 – 14
coefficients of dividend
= x2 – 3x + 6 3
x– 14
Insert zero coefficient as placeholder for the missing x2 term.
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Remainder Theorem: The remainder of the division of a polynomial f (x) by x – a is f (a).
Example: Using the remainder theorem, evaluate f(x) = x
4 – 4x – 1 when x = 3.
91 0 0 – 4 – 13
1
3
3 9
6927
23 68
The remainder is 68 at x = 3, so f (3) = 68.
You can check this using substitution: f(3) = (3)4 – 4(3) – 1 = 68.
value of x
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Example: Using synthetic division and the remainder theorem, evaluate f (x) = x2 – x at x = – 2.
6
1 – 1 0– 2
1
– 2
– 3 6
Then f (– 2) = 6 and (– 2, 6) is a point on the graph of f(x) = x2 – x.
f(x) = x2 – x
x
y
2
4
(– 2, 6)
remainder
