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Division of polynomials concept is very similar to the normal division. As a polynomial contains a long string of monomials, each of the monomials is taken into consideration while dividing it. The long division method is the most commonly used division method for polynomials. The number of monomials (terms) in a polynomial of the dividend and divisor might differ but they all follow a similar division method.
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Key Takeaways - Monomial, Binomial, Polynomial, Long Division Method, Degree of Polynomial, Remainder theorem, Roots of Polynomial
Read Also: Multiplying a Monomial by a Polynomial
Order and Division of Polynomials
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Polynomial is an algebraic expression of the type
anxn + an−1xn−1+…………………a2x2 + a1x + a0
where “n” is either 0 or positive variables and real coefficients.
It has more than two terms. It is important to arrange the terms of a polynomial according to their ‘order’ before division. The higher-order terms are placed first followed by other terms ascendingly. For eg. in the polynomial x2 + 2x – 7, the term x2 has the highest order of ‘2’ followed by 2x with the order ‘1’ and 7 with the order ‘0’.

Terms of Polynomial
The order of the term in a polynomial is decided based on the sum of the highest power to which x (unknown variable) is raised. There are many methods of division for polynomials, but the long division method is applicable for all types of polynomials and is simpler to understand. Concisely, it deals with the division of every term of the divisor with every term of the dividend. Based on the number of terms, an expression is named as monomial, binomial, and polynomial.
The video below explains this:
Polynomials Detailed Video Explanation:
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Types of Polynomial Division
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The different types for polynomial division are given below:
Division of a monomial with a monomial
A monomial is an expression which has only one term. It has the simplest division. Let us consider 60x2 divided by 10x,
\(\frac{60x^2}{10x} = \frac{6×10×x×x}{1×10×x} = 6x\)
Division of a polynomial with a monomial
A polynomial contains more than two terms. Each term of the polynomial is treated separately and divided individually with the monomial.
For eg. Divide 24x3 – 12xy + 9x by 3x
Firstly, we need to check if the polynomial is arranged in the ascending order of its terms. 24x3 has the order 3, -12xy has the order 2 (adding both the exponents of the unknown terms x1 and y1), 9x has the order 1. Thus, the polynomial is properly arranged. Here the dividend is the polynomial 24x3 – 12xy + 9x and the divisor is the monomial 3x. Now using the long division method,
\(\frac{24x^3-12xy+9x}{3x}\)
Now divide every term of the dividend with the monomial 3x,
8x2 – 4y +3
| 3x | 24x3 – 12xy + 9x |
| - | 24x3 – 12xy + 9x |
| 0 |
We need to find such a term for the quotient which when multiplied with the divisor will give the dividend’s each term as the answer. The answer is the quotient 8x2 – 4y +3.
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Division of a Polynomial with a Binomial
It is important that both the binomial and the polynomial are arranged in the order of their terms before division. A binomial having two terms, each of the terms is divided individually with the polynomial. The binomial is the divisor, and the polynomial is the dividend.
For eg. Divide 3x3 – 8x + 5 with x – 1
The polynomial 3x3 – 8x + 5 has 3,1,0 as the order of its terms. There is a missing order which can be written as 0x2 to complete the ascending order of the polynomial. Thus the polynomial is 3x3 + 0x2 – 8x + 5. The binomial x - 1 has the order 1,0.
3x2
| x – 1 | 3x3 + 0x2 – 8x + 5 |
| - | 3x2 - 3x2 |
| 0 + 3x2 – 8x + 5 |
Here, the first term of the binomial ‘x’ is used to divide the first term of the polynomial. The term 3x2 is written in the quotient because we know that by multiplying the first term of the binomial ‘x’ with ‘3x2’ will give us the first term of the polynomial ‘3x3’. After getting the quotient ‘3x2’ the second term of the binomial ‘-1’ is multiplied with the same quotient ‘3x2’ and its answer ‘-3x2’ is written next to the previous answer. By subtracting, all the signs of the terms change, and the remaining terms of the polynomials are carried forward for further division.
3x2 + 3x - 5
| x – 1 | 3x3 + 0x2 – 8x + 5 |
| - | 3x2 - 3x2 |
| + 3x2 – 8x + 5 | |
| - | 3x2 – 3x |
| 0 – 5x + 5 | |
| - | – 5x + 5 |
| 0 |
The division continues further with an aim to eliminate all the terms by adding apt terms to the quotient.
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Division of a polynomial with a polynomial
This type of division proceeds very similar to the previous type, the only difference is that the divisor will now have more than two terms in its expression.
For eg. Divide x2 + 2x + 3x3 + 5 by 1 + 2x + x2
Arranging the dividend x2 + 2x + 3x3 + 5 in its order → 3x3 + x2 + 2x + 5
Arranging the divisor 1 + 2x + x2 in its order » x2 + 2x + 1
Obtain the first term of the quotient in such a way that multiplying it with the first term of the divisor will give the first term of the dividend as the answer in order to eliminate it through subtraction. Continue to multiply the same term of the quotient with the remaining terms of the divisor and fill in the answers below the terms of the dividend of a similar order.
3x - 5
| x2 + 2x + 1 | 3x3 + x2 + 2x + 5 |
| - | 3x3 + 6x2 + 3x |
| - 5x2 – x + 5 | |
| - | - 5x2 – 10x - 5 |
| 0 + 9x + 10 |
The division cannot continue further as the degree of the remainder is less than the degree of the divisor.
Read Also: Permutations and Combinations
Remainder Theorem
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While dividing 15 by 6, we get 2 as the quotient and 3 as the remainder. So, in an equation form it is expressed as 15 = (6 × 2) + 3 as per the formula –
Dividend = (Divisor × Quotient) + Remainder
The Remainder theorem in polynomials helps in finding the remainder of a polynomial without actually dividing it. Let us consider p(x) as the dividend, x – a as the divisor, q(x) as quotient, and r(x) as remainder. Using the above given formula,
p(x) = (x - a) × q(x) + r(x)
The remainder is always constant so we can also write as,
p(x) = (x - a) × q(x) + r
Substituting x as a,
p(a) = (a - a) × q(a) + r
p(a) = (0) × q(a) + r
p(a) = r
Hence it is proved that we can find the remainder of any division if we know the ‘a’ term of the divisor.
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Things to Remember
- Ensure to set the terms of a polynomial in standard order before proceeding to divide it.
- If there is a missing order in the series of terms in a polynomial, one should add 0xn, where n is the value of the missing order.
- The degree of a polynomial is the value of the highest order of its terms.
- Long division method concisely deals with the division of every term of the divisor with every term of the dividend.
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Sample Questions
Ques. Find the degree of the polynomials given below – (3 marks)
(1) 17x4 + 5x3y + 2
(2) x2y3 + x2 + x4y2 + 3
(3) x + xy + 3x2 + 1
(4) a3b + a2 + 9
Ans. The highest order in the polynomial becomes the degree of a polynomial.
- The highest order in the polynomial is 4, thus the degree of polynomial is also 4.
- The third term has the highest order with a sum of 4 + 2 = 6 exponents; thus its degree will be 6.
- The degree of the polynomial is 2.
- The degree of the polynomial is 4.
Ques. Arrange the following polynomial in the ascending order of their exponents – (3 marks)
(1) x + 4x2 + 5
(2) y3 + 2 + y2 + y
(3) a2 + a + 4x3
(4) b2 + 6b + 8b3 + 4 + b5
Ans.
- 4x2 + x + 5
- y3 + y2 + y + 2
- 4x3 + a2 + a + 0
- b5 + 0b4 + 8b3 + b2 + 6b + 0
Ques. Divide x2 – 3x – 10 by x + 2 (3 marks)
Ans. x - 5
| x + 2 | x2 – 3x – 10 |
| - | x2 + 2x |
| 0 – 5x - 10 | |
| - | - 5x - 10 |
| 0 |
Ques. Divide 2x2 – 5x – 1 by x - 3 (3 marks)
Ans. 2x - 5
| x- 3 | 2x2 – 5x – 1 |
| - | 2x2 - 6x |
| 0 + x – 1 | |
| - | x - 3 |
| 0 + 2 |
Ques. Divide x6 + 2x4 + 6x - 9 by x3 + 3 (3 marks)
Ans. x3 + 2x - 3
| x3 + 3 | x6 + 0x5 + 2x4 + 0x3 + 0x2 + 6x - 9 |
| - | x6 + 3x3 |
| 0 2x4 - 3x3 + 6x - 9 | |
| - | 2x4 + 6x |
| -3x3 - 9 | |
| - | - 3x3 - 9 |
| 0 |
Ques. Divide x2 + 2x2y - 2xy + 2xy2 – 3y2 by x + y (3 marks)
Ans. 2xy + x – 3y
| x + y | 2x2y + 2xy2 + x2 – 2xy – 3y2 |
| - | 2x2y + 2xy2 |
| 0 + x2 – 2xy – 3y2 | |
| - | x2 + xy |
| 3xy – 3y2 | |
| - | - 3xy – 3y2 |
| 0 |
Ques. Check whether the polynomial q(t) = 4t3 + 4t2 – t – 1 is a multiple of 2t + 1. (3 marks)
Ans. q(t) will be a multiple of 2t + 1 only, if 2t + 1 divides q(t)leaving the remainder zero. Now, taking 2t + 1 = 0, we have t = -\(\frac{1}{2}\).
So the remainder obtained on dividing q(t) by 2t + 1 is 0.
Thus, 2t + 1 is a factor of the given polynomial q(t), that is q(t) is a multiple of 2t + 1.
Ques. Solve (x4 - 10x3 + 27x2 - 46x + 28) ÷ (x - 7). (3 marks)
Ans.
Ques. Solve (4x3 + 5x2 + 5x + 8) ÷ (4x + 1). (3 marks)
Ans.
Ques. Find the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1. (2 marks)
Ans. Here, p(x) = x4 + x3 – 2x2 + x + 1, and the zero of x – 1 is 1.
So, p (1) = (1)4 + (1)3 – 2(1)2 + 1 + 1
p (1) = 2
So, by the Remainder Theorem, 2 is the remainder when x4 + x3 – 2x2 + x + 1 is divided by x – 1.
Ques. Find the remainder obtained on dividing p(x) = x3 + 1 by x + 1. (2 marks)
Ans.
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