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    You are at:Home»Pure Math»Algebra»Dividing Polynomials | Division of polynomials examples with solutions
    Dividing polynomials long division examples with solutions | polynomial long division examples | Dividing Polynomials by monomials, trinomials & Quadratics

    Dividing Polynomials | Division of polynomials examples with solutions

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    By sivaalluri on November 27, 2018 Algebra

    Division of a polynomial by another polynomial is one of the important  concept in Polynomial expressions. In this article explained about  basic phenomena of diving polynomial algorithm in step by step process

    Table of Contents

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    • Algebra division| Dividing Polynomials Long Division
      • Polynomial long division examples with solution
        • Dividing polynomials by monomials
        • Dividing Polynomials by Binomials
        • Dividing Polynomials by Trinomials
        • Dividing Polynomials by Quadratics

    Algebra division| Dividing Polynomials Long Division

    Before going to algebra divisions observe the normal numerical division algorithm

    When we divide 137 by 5 we get the quotient 27 and remainder 2

    We can write 137 as

    137 = (5 x 27) + 2   (Note : Here remainder 2 and it is less than divisor 5)

    i.e Dividend = Divisor x Quotient + Remainder

    The same division algorithm of number is also applicable for division algorithm of polynomials.

    i.e When a polynomial divided by another polynomial

    Dividend = Divisor x Quotient + Remainder, when remainder is zero or polynomial of degree less than that of divisor

    A polynomial f(x) is divided by another polynomial g(x) we get quotient q(x) and remainder p(x) such that

    f(x) = g(x) . q(x) + p(x)

    Where p(x) = 0 or degree of p(x) < degree of g(x)

    Polynomial long division examples with solution

    Dividing polynomials by monomials

    Take one example

    Example -1 :  Divide the polynomial 2x4 +3x2 +x  by x

    Here   \frac{2x^4 + 3x^2 + x}{x}

    =  \frac{2x^4}{x} + \frac{3x^2}{x} + \frac{x}{x}

    = 2x3 + 3x +1

    So we write the polynomial 2x4 +3x2 +x as product of x and 2x3 + 3x +1

    2x4 +3x2 +x = (2x3 + 3x +1) x

    It means x & 2x3 + 3x +1 are factors of 2x4 +3x2 +x

    Example – 2 : Divide the polynomial 3x3 + 9x2 + 5 by 3x

    = \frac{3x^3 + 9x^2 + 5}{3x}

    =   \frac{3x^3}{3x} + \frac{9x^2}{3x} + \frac{5}{3x}

     = 3x^2 + 3x + \frac{5}{3x}

    Here variable “x” in denominator so it is not a polynomial.

    So polynomial of 3x3 + 9x2 + 5 can be written as

    3x3 + 9x2 + 5 = 3x (x2 + 3x) +5

    Here we can say (x2 + 3x) is the quotient, 3x is the divisor and 5 is the remainder. Also we can say that the reminder is not zero, 3x is not a factor of 3x3 + 9x2 + 5

    Dividing Polynomials by Binomials

    Example – 3: Divide the polynomial 4x2-3x +x3 +10 by x+4 and verify the remainder with zero of the divisor

    Solution: Say f(x) = 4x2-3x +x3 +10  and g(x)= x+4

    For this division, we have using the steps as follows

    Step: 1 – Arrange the terms of the dividend f(x) and the divisor g(x) in descending order of their degree

    f(x)= x3 +4x2-3x  +10 and g(x) = x+4

    Step:2 – We get the first term of the quotient by dividing the first term of the dividend (i.e  x3 ) with the first term of the divisor (i.e x)

    So  = \frac{x^3}{x} = x^2

    Step:3 – Now the first term of quotient x2 multiplying the divisor g(x) and that term subtracted from the dividend

    x2 (x +4) = x3 +4x2

    Step : 4 –  Now repeat the step 2 again

    i.e  To get the second term of the quotient by dividing the first term of get remainder in the previous step (i.e  3x ) with the first term of the divisor (i.e x)

    So  = \frac{3x}{x} = 3

    Step:5 – Then the second term of quotient 3 is multiplied by the divisor g(x) and that term is subtracted from the remainder in step 3

    3(x +4) = 3x + 12

    Here the quotient is (x2 – 3) and remainder is 22

    Here we can write this division process as

    Dividend = Divisor x Quotient + Remainder

    f(x)  = (x2 – 3) g(x) + 22

    x3 + 4x2 – 3x  +10 =  (x2 – 3) ( x+4) + 22

    By using the remainder theorem we obtain the remainder  without performing the above process

    Note: Remainder theorem is applicable only for divisors that have linear polynomials with one variable

    Consider zero of the polynomial of g(x) is   – 4

    g(x) = x + 4

    g(-4) = -4 +4 = 0

    Now find the value of f(x) at x= -4

    f(4) = 4 (-4)2 – 3(-4) + (-4)3 + 10 = 22

    Dividing Polynomials by Trinomials

    Example  -4:  Divide the polynomial  5x5  + x2  + 3x3 –  4x4– 10  by  2 +  x2 – x

    Solution:  Say f(x) = 5x5  + x2  + 3x3 –  4x4– 10 and g(x)= 2 +  x2 – x

    Now divide f(x) by g(x) by the following steps

    Step: 1 – Arrange the terms of the dividend f(x) and the divisor g(x) in descending order of their degree

    f(x)= 5x5  – 4x4  + 3x3  + x2 – 10  and g(x) = x2 – x  + 2

    Step:2 – We get the first term of the quotient by dividing the first term of the dividend (i.e  5x5 ) with the first term of the divisor (i.e x2)

    So   = \frac{5x^5}{x^2} = 5x^3

    Step : 3 – Now the first term of quotient 5x3 multiplied by the divisor g(x) and that term is subtracted from the dividend

    5x3 (x2 – x  + 2) = 5x3 – 5x4 + 10x3

    Step : 4 –  Now repeat the step 2 again

    i.e  To get the second term of the quotient by dividing the first term of get remainder in the previous step (i.e  x4 ) with the first term of the divisor (i.e x2)

    So  = \frac{x^4}{x^2} = x^2

    Step:5 – Then the second term of quotient x2 is multiplied by the divisor g(x) and that term is subtracted from the remainder got in step 3

    x2 (x2 – x  + 2) = x4 – x3  + 2x2

    Step : 6 –  Now repeat the step 4 again

    i.e  To get the third term of the quotient by dividing the first term of get remainder in the previous step (i.e  -6x3 ) with the first term of the divisor (i.e x2)

    So  = \frac{-6x^3}{x^2} = -6x

    Step:7 – Then the third term of the quotient -6x multiplying the divisor g(x) and that term subtracted from the remainder got in step 5

    -6x (x2 – x  + 2) = -6x3  + 6x2  – 12x

    Step: 8 –  Now repeat the step 6 again

    i.e  To get the fourth term of the quotient by dividing the first term of the remainder in the previous step (i.e  -7x2 ) with the first term of the divisor (i.e x2)

    So  = \frac{-7x^2}{x^2} = -7

    Step:9 – Then the fourth term of quotient -7 is multiplied by the divisor g(x) and that term is subtracted from the remainder in step 7

    -7 (x2 – x  + 2) = -7x2  + 7x  – 14

    Here the division process is completed because the degree of (5x + 4)  = 1 which is less than the degree of the divisor (x2 – x  + 2)

    So the quotient is (5x3  + x3  -6x – 7) and remainder is 5x + 4

    Here we can write this division process as

    Dividend = Divisor x Quotient + Remainder

    f(x)  =  (5x3  + x3  -6x – 7) g(x) + ( 5x + 4)

    (5x5  + x2  + 3x3 –  4x4– 10)  =  (5x3  + x3  -6x – 7)  (x2 – x  + 2)  + ( 5x + 4)

    Dividing Polynomials by Quadratics

    Example: Divide the polynomial   1 + x  + 5x5  -4x4  + 3x3 – 2x2  by  2 + x +  x2  + x3

    Solution:  Say f(x) = 1 + x  + 5x5  -4x4  + 3x3 – 2x2  and g(x)= 2 + x +  x2  + x3

    Now divide f(x) by g(x) by the following steps

    Step: 1 – Arrange the terms of the dividend f(x) and the divisor g(x) in descending order of their degree

    f(x)= 5x5  -4x4  + 3x3 – 2x2 + x +1   and g(x) = x3 +  x2  + x + 2

    Step:2 – We get the first term of the quotient by dividing the first term of the dividend (i.e  5x5 ) with the first term of the divisor (i.e x3)

    So   = \frac{ 5x^5}{x^3} = 5x^2

    Step:3 – Now the first term of quotient 5x2 multiplying the divisor g(x) and that term subtracted from dividend

    5x2 (x3 +  x2  + x + 2) = 5x5 + 5x4 + 5x3  +  10x2

    Step: 4 –  Now repeat the step 2 again

    i.e  To get the second term of the quotient by dividing the first term of get remainder in the previous step (i.e  -9x4 ) with the first term of the divisor (i.e x3)

    So   = \frac{-9x^4}{x^3} = -9x

    Step:5 – Then the second term of the quotient -9x multiplying the divisor g(x) and that term subtracted from the remainder got in step 3

    -9x  (x3 +  x2  + x + 2)  = -9x4 – 9x3  – 9x2  – 18x

    Step: 6 –  Now repeat the step 4 again

    i.e  To get the third term of the quotient by dividing the first term of get remainder in the previous step (i.e  7x3 ) with the first term of the divisor (i.e x3)

    So  = \frac{7x^3}{x^3} = 7

    Step:7 – Then the third term of quotient  7 is multiplied by the divisor g(x) and that term is subtracted from the remainder got in step 5

    7 (x3 +  x2  + x + 2) = 7x3 + 7 x2  + 7x + 14

    Here the division process is completed because the degree of (- 10x2 + 12x  -13 )  = 2 which is less than the degree of the divisor (x3 +  x2  + x + 2)

    So the quotient is (5x2  -9x  + 7) and remainder is  -10x2  + 12x  -13

    Here we can write this division process as

    Dividend = Divisor x Quotient + Remainder

    f(x)  =   (5x2  -9x  + 7) g(x) + ( -10x2  + 12x  -13 )

    (5x5  -4x4  + 3x3 – 2x2 + x +1) =   (5x2  -9x  + 7) (x3 +  x2  + x + 2) + ( -10x2  + 12x  -13 )

    Dividing polynomials long division examples with solutions | polynomial long division examples | Dividing Polynomials by monomials, trinomials & Quadratics

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    Related Topics

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    Remainder and Factor Theorem Proof | Remainder and Factor Theorem Test

    Remainder Theorem Tough Questions for Competitive Exams | Aptitude Questions

    Algebra Divide the polynomial dividing polynomials division polynomial
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