Dividing Polynomials Examples Using Long Division
This page provides dividing polynomials examples and problems, using long division.
You may wish to view the examples on dividing polynomials below the exercises before attempting to answer the questions.
Dividing Polynomials – Exercises
Instructions: Perform the division as shown in each problem.
1) (3x – 12) ÷ 3 = ?
2) (16x3 + 8x2 + 4x) ÷ (2x) = ?
3) (84a2b + 16ab2 + 36ab – 12ab3) ÷ (4ab) = ?
4) (15a2 + 22a + 8) ÷ (3a + 2) = ?
5) (x2 – 2x + 5) ÷ (x – 1) = ?
Dividing Polynomials – Answers
1) The correct answer is: x – 4
(3x – 12) ÷ 3 =
(3x ÷ 3) – (12 ÷ 3) =
(3x ÷ 3) – (4) =
x – 4
2) The correct answer is: 8x2 + 4x + 2
(16x3 + 8x2 + 4x) ÷ (2x) =
(16x3 ÷ 2x) + (8x2 ÷ 2x) + (4x ÷ 2x) =
(16x3 ÷ 2x) + (8x2 ÷ 2x) + (2) =
(16x3 ÷ 2x) + (4x) + 2 =
8x2 + 4x + 2
3) The correct answer is: 21a – 3b2 + 4b + 9
(84a2b + 16ab2 + 36ab – 12ab3) ÷ (4ab) =
(84a2b ÷ 4ab) + (16ab2 ÷ 4ab) + (36ab ÷ 4ab) – (12ab3 ÷ 4ab) =
(84a2b ÷ 4ab) + (16ab2 ÷ 4ab) + (36ab ÷ 4ab) – (3b2) =
(84a2b ÷ 4ab) + (16ab2 ÷ 4ab) + (9) – 3b2 =
(84a2b ÷ 4ab) + (16ab2 ÷ 4ab) + 9 – 3b2 =
(84a2b ÷ 4ab) + (4b) + 9 – 3b2 =
(21a) + 4b + 9 – 3b2 =
21a – 3b2 + 4b + 9
4) The correct answer is: 5a + 4
5a + 43a + 2) 15a2 + 22a + 8
–(15a2 + 10a)
12a + 8
–(12a + 8)
0
5) The correct answer is: \(\text x – 1 + \frac{4}{(x – 1)}\)
x – 1x– 1) x2 – 2x + 5
–(x2 – x)
–x + 5
–(–x + 1)
4
As shown in the solution to problem 5, solutions to questions on dividing polynomials can sometimes have a remainder.
To see more free math sample questions, please click here.
To learn how to divide polynomial equations using factoring or synthetic division, please see our posts on:
Dividing Polynomials by Monomials – Example
A monomial is a single algebraic term.
In other words, a monomial is not an expression, so it won’t have the plus or minus sign.
A monomial can be a number or a number with a variable.
Divide a polynomial by a monomial using following steps.
Example: (6x2 – 2x) ÷ 2x = ?
Answer: 3x – 1
STEP 1: Expand the problem using sets of parentheses.
(6x2 – 2x) ÷ 2x =
(6x2 ÷ 2x) – (2x ÷ 2x)
STEP 2: Find the factors for the terms in each set of parentheses.
3x × 2x = 6x2 so (6x2 ÷ 2x) = 3x
1 × 2x = 2x so (2x ÷ 2x) = 1
STEP 3: Take the equation from step 1 and simplify for each set of parentheses.
(6x2 ÷ 2x) – (2x ÷ 2x) =
(6x2 ÷ 2x) – 1 =
3x – 1
Dividing Polynomials by Binomials – Examples
To divide a polynomial by a binomial or by another polynomial, you can use long division.
Example: (m3 – m) ÷ (m + 1) = ?
Answer: m2 – m
STEP 1: Set up the long division. Be sure to put in the missing terms.
For example, if you have a polynomial with m3 but not m2 , like this example, then you can put in placeholders.
To use placeholders, be sure that your dividend has all of the variables with the exponents in descending order, and put in zeros where necessary.
m + 1) m3 + 0m2 – m
STEP 2: Divide the first term from the divisor (m) into the first term of the dividend (m3 ) using long division.
According to the laws of exponents we know that:
m3 ÷ m = m2
So, use m2 as the first term in the quotient.
m2
m + 1) m3 + 0m2 – m
STEP 3: Then multiply the first term that you put in the quotient (m2) by the divisor (m + 1) and write the result down below the divisor.
m2
m + 1) m3 + 0m2 – m
–(m3 + m2)
STEP 4: Now subtract the result of step 3 from the dividend.
m2
m + 1) m3 + 0m2 – m
–(m3 + m2)
–m2
STEP 5: Bring down the next term from the dividend (–m).
m2
m + 1) m3 + 0m2 – m
–(m3 + m2)
–m2 – m
STEP 6: Divide the first term from the divisor (m) into the first term from step 5 (–m2).
–m2 ÷ m = –m
Place this result (–m) in the quotient at the top of the problem.
m2 – m
m + 1) m3 + 0m2 – m
–(m3 + m2)
–m2 – m
STEP 7: Then multiply the second number in the quotient (–m) by the divisor (m + 1) and write the result down below.
m2 – m
m + 1) m3 + 0m2 – m
–(m3 + m2)
–m2 – m
–(–m2– m)
STEP 8: Now subtract to get your result.
m2 – m
m + 1) m3 + 0m2 – m
–(m3 + m2)
–m2 – m
–(–m2 – m)
0
For an alternative to solve these types of questions on polynomial division, please see our post entitled: Remainder Theorem
Dividing Polynomials – Quadratics
You might see problems on polynomials in the quadratic form on your algebra test.
These polynomial division problems have equations in one of the following forms:
ax2 ± bx ± c = 0
x2 ± bx ± c = 0
x2 ± bx = n
x2 = n
To learn how to solve polynomial equations that are in the quadratic form, please see our posts on:
Problems on dividing polynomials can also solved by setting up the polynomial expressions as a fraction.
You may also want to have a look at our post on algebraic fractions.
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