Long Division of Polynomials
Lori Jordan
Kate Dirga
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Printed: November 22, 2013
AUTHORS
Lori Jordan
Kate Dirga
www.ck12.org
C ONCEPT
Concept 1. Long Division of Polynomials
1
Long Division of
Polynomials
Here you’ll learn how to use long division to divide polynomials.
The area of a rectangle is 6x3 − 12x2 + 4x − 8. The width of the rectangle is 2x − 4. What is the length?
Watch This
MEDIA
Click image to the left for more content.
Khan Academy: Polynomial Division
Guidance
Even though it does not seem like it, factoring is a form of division. Each factor goes into the larger polynomial
evenly, without a remainder. For example, take the polynomial 2x3 − 3x2 − 8x + 12. If we use factoring by grouping,
we find that the factors are (2x − 3)(x − 2)(x + 2). If we multiply these three factors together, we will get the original
polynomial. So, if we divide by 2x − 3, we should get x2 − 4.
2x − 3 ) 2x3 −3x2 −8x +12
How many times does 2x go into 2x3 ? x2 times.
x2
2x − 3 ) 2x3 −3x2 −8x + 12
2x3 −3x2
0
Place x2 above the x2 term in the polynomial.
Multiply x2 by both terms in the divisor (2x and -3) and place them until their like terms. Subtract from the dividend
(2x3 − 3x2 − 8x + 12). Pull down the next two terms and repeat.
x2
−4
3
2
2x − 3 ) 2x −3x −8x +12
2x3 −3x2
−8x +12
−8x +12
1
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2x goes into −8x 4 times.
After multiplying both terms in the divisor by -4, place that under the terms you brought down. When subtracting
we notice that everything cancels out. Therefore, just like we thought, x2 − 4 is a factor.
When dividing polynomials, not every divisor will go in evenly to the dividend. If there is a remainder, write it as a
fraction over the divisor.
Example A
(2x3 − 6x2 + 5x − 20) ÷ (x2 − 5)
Solution: Set up the problem using a long division bar.
x2 − 5 ) 2x3 −6x2 +5x −20
How many times does x2 go into 2x3 ? 2x times.
x2 − 5
2x
) 2x3 −6x2 +5x −20
2x3 −10x2
4x2 +5x −20
Multiply 2x by the divisor. Subtract that from the dividend.
Repeat the previous steps. Now, how many times does x2 go into 4x2 ? 4 times.
x2 − 5
2x +4
) 2x3 −6x2 +5x −20
2x3 −10x2
4x2 +5x −20
4x2
−20
5x
At this point, we are done. x2 cannot go into 5x because it has a higher degree. Therefore, 5x is a remainder. The
complete answer would be 2x + 4 + x25x−5 .
Example B
(3x4 + x3 − 17x2 + 19x − 6) ÷ (x2 − 2x + 1). Determine if x2 − 2x + 1 goes evenly into 3x4 + x3 − 17x2 + 19x − 6. If
so, try to factor the divisor and quotient further.
Solution: First, do the long division. If x2 − 2x + 1 goes in evenly, then the remainder will be zero.
2
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Concept 1. Long Division of Polynomials
3x2 +7x −6
x2 − 2x + 1 ) 3x4 +x3 −17x2 +19x −6
3x4 −6x3 3x2
7x3 −20x2 +19x
7x3 −14x2 +7x
−6x2 +12x −6
−6x2 +12x −6
0
This means that x2 − 2x + 1 and 3x2 + 7x − 6 both go evenly into 3x4 + x3 − 17x2 + 19x − 6. Let’s see if we can factor
either x2 − 2x + 1 or 3x2 + 7x − 6 further.
x2 − 2x + 1 = (x − 1)(x − 1) and 3x2 + 7x − 6 = (3x − 2)(x + 3).
Therefore, 3x4 + x3 − 17x2 + 19x − 6 = (x − 1)(x − 1)(x + 3)(3x − 2). You can multiply these to check the work.
A binomial with a degree of one is a factor of a larger polynomial, f (x), if it goes evenly into it. In this example,
(x − 1)(x − 1)(x + 3) and (3x − 2) are all factors of 3x4 + x3 − 17x2 + 19x − 6. We can also say that 1, 1, -3, and 23
are all solutions of 3x4 + x3 − 17x2 + 19x − 6.
Factor Theorem: A polynomial, f (x), has a factor, (x − k), if and only if f (k) = 0.
In other words, if k is a solution or a zero, then the factor, (x − k) divides evenly into f (x).
Example C
Determine if 5 is a solution of x3 + 6x2 − 8x + 15.
Solution: To see if 5 is a solution, we need to divide the factor into x3 + 6x2 − 8x + 15. The factor that corresponds
with 5 is (x − 5).
x2 +11x +5
x − 5 ) x3 +6x2 −50x +15
x3 −5x2
11x2 −50x
11x2 −55x
5x +15
5x −25
40
Since there is a remainder, 5 is not a solution.
Intro Problem Revisit
First, do the long division.
3x2
+2
3
2
2x − 4 ) 6x −12x +4x −8
6x3 −12x2
4x −8
4x −8
0
3
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This means that 2x − 4 and 3x2 + 2 both go evenly into 6x3 − 12x2 + 4x − 8.
3x2 + 2 can’t be factored further, so it is the rectangle’s length.
Guided Practice
1. (5x4 + 6x3 − 12x2 − 3) ÷ (x2 + 3)
2. Is (x + 4) a factor of x3 − 2x2 − 51x − 108? If so, find any other factors.
3. What are the real-number solutions to #2?
4. Determine if 6 is a solution to 2x3 − 9x2 − 12x − 24.
Answers
1. Make sure to put a placeholder in for the x−term.
5x2 +6x −27
x2 + 3 ) 5x4 +6x3 −12x2 +0x −3
5x4
+15x2
6x3 −27x2 +0x
6x3
+18x
2
−27x −18x −3
−27x2
−81
−18x +78
The final answer is 5x2 + 6x − 27 − 18x−78
.
x2 +3
2. Divide (x + 4) into x3 − 2x2 − 51x − 108 and if the remainder is zero, it is a factor.
x2 −6x −27
x + 4 ) x3 −2x2 −51x −108
x3 +4x2
−6x2 −51x
−6x2 −24x
−27x −108
−27x −108
0
x + 4 is a factor. Let’s see if x2 − 6x − 27 factors further. Yes, the factors of -27 that add up to -6 are -9 and 3.
Therefore, the factors of x3 − 2x2 − 51x − 108 are (x + 4), (x − 9), and (x + 3).
3. The solutions would be -4, 9, and 3; the opposite sign of each factor.
4. To see if 6 is a solution, we need to divide (x − 6) into 2x3 − 9x2 − 12x − 24.
4
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Concept 1. Long Division of Polynomials
2x2 +3x +6
x − 6 ) 2x3 −9x2 −12x −24
2x3 −12x2
3x2 −12x
3x2 −18x
6x −24
6x −36
12
Because the remainder is not zero, 6 is not a solution.
Vocabulary
Long division (of polynomial)
The process of dividing polynomials where the divisor has two or more terms.
Divisor
The polynomial that divides into another polynomial.
Dividend
The polynomial that the divisor goes into. The polynomial under the division bar.
Quotient
The answer to a division problem.
Practice
Divide the following polynomials using long division.
1.
2.
3.
4.
5.
6.
(2x3 + 5x2 − 7x − 6) ÷ (x + 1)
(x4 − 10x3 + 15x − 30) ÷ (x − 5)
(2x4 − 8x3 + 4x2 − 11x − 1) ÷ (x2 − 1)
(3x3 − 4x2 + 5x − 2) ÷ (3x + 2)
(3x4 − 5x3 − 21x2 − 30x + 8) ÷ (x − 4)
(2x5 − 5x3 + 6x2 − 15x + 20) ÷ (2x2 + 3)
Determine all the real-number solutions to the following polynomials, given one factor.
7. x3 − 9x2 + 27x − 15; (x + 5)
8. x3 + 4x2 − 9x − 36; (x + 4)
9. 2x3 + 7x2 − 7x − 30; (x − 2)
Determine all the real number solutions to the following polynomials, given one zero.
10. 6x3 − 37x2 + 5x + 6; 6
11. 6x3 − 41x2 + 58x − 15; 5
12. x3 + x2 − 16x − 16; 4
5
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Find the equation of a polynomial with the given zeros.
13.
14.
15.
16.
6
4, -2, and 23
1, 0, and 3
-5, -1, and 34
Challenge Find two polynomials with the zeros 8, 5, 1, and -1.