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    You are at:Home»Pure Math»Algebra»Factor theorem for polynomial | Factoring polynomials using factor theorem
    Factorise the Polynomial by using Factor Theorem | allmathtricks
    Algebra

    Factor theorem for polynomial | Factoring polynomials using factor theorem

    sivaalluriBy sivaalluriDecember 7, 2018Updated:February 12, 2025No Comments8 Mins Read
    If a polynomial f(x) is dividing by g(x) leaves remainder zero then g(x) is a factor of f(x). Here explained about factor theorem example with solution of factorise the polynomials by using factor theorem

    Factor Theorem Applications| Factor theorem example problems

    Factor theorem state and proof purpose go through the below link

    Statement and proof of Factor Theorem

    Factor theorem practice problems

    Example – 1 : Find the value of k, if (2x-3) is a factor of 2x3 – 9x2 + x + k

    Solution: Let f(x) = 2x3 – 9x2 + x + k  &  g(x) = 2x-3

    Since  g(x) is a factor of f(x)

    This implies that f(3/2) = 0

    Now put x = 3/2 in f(x)

    -48/4 + K = 0

    K = 12

    Example – 2 :Find the value of “n” and ‘m” such that the polynomial x3 – nx2– 13x + m has (x  + 1) and ( x – 2) as factor.

    Solution: Let f(x) =x3 – nx2– 13x + m

    Since (x +1) is a factor of  f(x)

    i.e f(-1) = 0

    So f ( -1) =  – 1 -n +13 +m = 0

    m -n = -12   – – – – – – – –  ( 1)

    Since (x -2) is a factor of  f(x)

    i.e f(2) = 0

    So f ( 3) = 8 -4n  – 26 + m = 0

    m -4n = 18   – – – – – – – – – (2)

    Now from equations (1) and (2)

    n = -10  & m = -22

    Example – 3 : What much be added to x3 – 3x2 + 4x – 15 to obtain a polynomial which is exactly divisible by  x-3 ?

    Solution: Let f (x) =  x3 – 3x2 + 4x – 15  &  g(x) = x -3

    Since the degree of the g(x) is 1. So the degree of the remainder is 0

    Let “b” be added to f(x), so that it may be exactly divisible by g(x)

    i.e f(x) = (x3 – 3x2 + 4x -15) + b

    Since f(x) is exactly divisible by (x-3)

    So f(3) = 0

    27 – 27 + 12 – 15 + b = 0

    b = 3

    i. e g(x) is a factor of f(x) while add constant of 3 to f(x)

    Application of the factor theorem

    Application 1  

    Prove that (x- a) is a factor of xn – an for any natural number of “n” by using factor theorem

    Let f(x) = xn – an   , g(x) = x –a Where n = Natural number

    The remainder when f(x) is divide by g(x) is f(a)

     Now put x = a in f(x) then

    f(a) = an – an   = 0

    So f(a) = 0 for any natural number of “n”

    i.e (x- a) is a factor of xn – an   where n \in N

    Application 2

    Prove that (x + a) is a factor of xn + an for any odd number of “n” by using factor theorem

    Let f(x) = xn + an   , g(x) = x + a Where n = odd number

    The remainder when f(x) is divide by g(x) is f(-a)

     Now put x = – a in f(x) then

    f(-a) = (-a)n + an   = – an + an   = 0

    So f(-a) = 0 for any  odd number of “n”

    i.e (x + a) is a factor of xn + an   where n  = odd number

    Example – 4 : Find the factors of the following polynomials

    i) a3 +8           ii) a6 – b6             iii) 125a3 + 216b3

    Solutions :

    i )  a3 +8     = a3 + 23    where exponential is odd number

      According to Application 2

    (a +2 ) is factor of  a3 + 23

    ii )  a6 – b6      where exponential is natural number

      According to Application 1

    (a – b ) is factor of  a6 – b6

    iii )  125a3 + 216b3    = (5a)3 + (6b)3  where exponential is odd number

      According to Application 2

    (5a + 6b ) is a factor of  125a3 + 216b3

    Application 3 (Factorisation of a quatratic polynomial)

    Factorise any quadratic polynomial ax2 + bx + c ( where a \neq 0 and a,b & c are constants ), we have to write b as the sum of two numbers whose product is “ac”.

    Proof:

    Here a polynomial ax2 + bx + c ( where a ≠ 0 and a,b & c are constants )

    Let its factors be   (rx + s) &  (px + q)  then

    ax2 + bx + c = (rx + s) (px + q)  = rpx2 + (sp + rq)x + sq.

    Comparing the coefficients of x2, we get that a = rp

    Similarly comparing the coefficient of x, we get b = sp + rq

    And when comparing the constant term, we get c = sq.

    This shows that “b” is the sum of two numbers sp & rq, whose product is

    (sp) (rp) =(rp)  (sq)  = a c

    Therefore, to factorise  ax2 + bx + c , we have to write ‘ b ‘ as the sum of two numbers whose product is ‘ ac ‘

    Example – 5 : Factorise using factor theorem 6x2 + 17x +5

    Solution: By using the application -3

    If we can find two numbers p and q such that p +q = 17 and pq =  6 x 5 = 30, we can get the factors

    So, let us look for the factors of 30. Some are

    1 & 30

    3 & 10

    2 & 15

    Of these pairs 2 & 15 will give us  p +q = 2 + 15 = 17

    So, 6x2 + 17x +5  = 6x2 + (15 +2)x +5

    = 2x(3x +1) + 5(3x + 1)

    = (2x +5) (3x +1)

    Therefore (2x +5) & (3x +1) are factors of a polynomial  6x2 + 17x +5

    Example – 6 : Factorise using factor theorem 2x2 + 5x + 3

    Solution: By using the application -3

    If we can find two numbers p and q such that p +q = 5 and pq =  2 x 3 = 6, we can get the factors

    So, let us look for the factors of 6. Some are

    1 & 6

    2 & 3

    Of these pairs 2 & 3 will give us  p +q = 2 +3 = 5

    So, 2x2 + 5x + 3 = 2x2 + (2+3)x + 3

    = x(2x +3) + (2x + 3)

    = (2x +3) (x +1)

    Therefore (2x +3) &  (x +1) are factors of a polynomial  2x2 + 5x + 3

    Example – 7 :  Using factor theorem factorize x4 +4x3 -x2 – 16x -12

    Solution: Let p(x) = x4 +4x3 -x2 – 16x -12

    The constant term of p(x) is  -12 and its factors are

    ±1 , ±2, ±3, ±4, ±6 & ±12

    So in beginning –  – –  – – –  till the value of p(x) is zero

    Put x = 1 in p(x)

    P(1) = 1 + 4 – 1 – 16 – 12 = 24 ≠ 0

    i.e (x – 1) is not a factor for p(x)

    Now put x = -1 in p(x)

    P(-1) = 1 -4 -1 +16 – 12 = 17 – 17 = 0

    i.e (x + 1) is a factor for p(x)

    By dividing p(x) with (x+1),

    We get the quotient x3 + 3 x2 – 4x – 12

    So p (x) = (x +1) (x3 + 3 x2 – 4x – 12)

    Let q(x) = x3 + 3 x2 – 4x – 12

    Again we should starts to find q(2), q(-2) –  –  –  –  –

    q(2) = 8 + 12 – 8 12 = 0

    Therefore ( x- 2) is a factor of q(x)

    By dividing q(x) with (x -2), we get quotient x2 + 5x + 6

    q(x) = (x -2) (x2 + 5x + 6)

    Finally

    P(x) = x4 +4x3 -x2 – 16x -12

    =  (x +1) (x3 + 3 x2 – 4x – 12)

    = (x +1) (x -2) (x2 + 5x + 6)

    = (x +1) (x -2) (x2 + 3 x +2x + 6)

    = (x +1) (x -2) [ x(x + 3) + 2(x + 3)]

    = (x +1) (x +2) (x -2)  (x + 3)

    Factorise the Polynomial by using Factor Theorem | allmathtricks

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