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Digital Lesson
Division of Polynomials
Dividing Polynomials
Long division of polynomials is similar to long division of
whole numbers.
When you divide two polynomials you can check the answer
using the following:
dividend = (quotient • divisor) + remainder
The result is written in the form:
remainder
dividend  divisor  quotient +
divisor
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Example: Divide x2 + 3x – 2 by x – 1 and check the answer.
x + 2
x  1 x 2  3x  2
x2 +
x
2x – 2
2x + 2
–4
remainder
Answer: x + 2 +
x2
1. x x   x
x
2. x( x  1)  x 2  x
2
3. ( x 2  3x)  ( x 2  x)  2 x
2x
4. x 2 x 
2
x
5. 2( x  1)  2 x  2
6. (2 x  2)  (2 x  2)  4
–4
x 1
Check: (x + 2) (x + 1) + (– 4) = x2 + 3x – 2 correct
quotient divisor remainder
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dividend
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Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer.
x2 + x + 3
2x  2 2x  0x  4x  1
3
2
2x3 – 2x2
2x2 + 4x
2x2 – 2x
6x – 1
6x – 6
5
Answer: x2 + x + 3 
5
2x  2
Check: (x2 + x + 3)(2x – 2) + 5
= 4x + 2x3 – 1
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Write the terms of the dividend in
descending order.
Since there is no x2 term in the
dividend, add 0x2 as a placeholder.
3
2
x
1.
2. x 2 (2 x  2)  2 x3  2 x 2
 x2
2x
2x2
3
3
2
2
3. 2 x  (2 x  2 x )  2 x
4.
x
2x
5. x(2 x  2)  2 x 2  2 x
6. (2 x 2  4 x)  (2 x 2  2 x)  6 x
8. 3(2 x  2)  6 x  6
7. 6 x  3
2x
9. (6 x  1)  (6 x  6)  5  remainder
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Example: Divide x2 – 5x + 6 by x – 2.
x – 3
x  2 x 2  5x  6
x2 – 2x
– 3x + 6
– 3x + 6
0
Answer: x – 3 with no remainder.
Check: (x – 2)(x – 3) = x2 – 5x + 6
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Example: Divide x3 + 3x2 – 2x + 2 by x + 3 and check the answer.
x2 + 0x – 2
x  3 x 3  3x 2  2 x  2
Note: the first subtraction
eliminated two terms from
the dividend.
Therefore, the quotient
skips a term.
Answer:
x2
–2+
8
x3
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x3 + 3x2
0x2 – 2x + 2
– 2x – 6
8
Check: (x + 3)(x2 – 2) + 8
= x3 + 3x2 – 2x + 2
6
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.
Since the divisor is x – 2, a = 2.
coefficients of the dividend
value of a
2
3
3
2
–1
6
16
8
15
coefficients of quotient
Answer: 3x + 8 
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1. Bring down 3
2. (2 • 3) = 6
3. (2 + 6) = 8
4. (2 • 8) = 16
5. (–1 + 16) = 15
remainder
15
x2
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Remainder Theorem: The remainder of the division of a
polynomial f (x) by x – k is f (k).
Example: Using the remainder theorem, evaluate
f(x) = x 4 – 4x – 1 when x = 3.
value of x
3
1
1
0
0
–4
–1
3
9
27
69
3
9
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.
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Factor Theorem: A polynomial f(x) has a factor (x – k) if and
only if f(k) = 0.
Example: Show that (x + 2) and (x – 1) are factors of
f(x) = 2x 3 + x2 – 5x + 2.
–2
2
2
1
–5
2
–4
6
–2
–3
1
0
1
2
2
–3
1
2
–1
–1
0
The remainders of 0 indicate that (x + 2) and (x – 1) are factors.
The complete factorization of f is (x + 2)(x – 1)(2x – 1).
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