Math, factoring

posted by Adeline

Two of the roots of x^4 +ax^3 +ax^2+11x+b=0 are 3 and -2.

1. Find the value of a and b

2. Find the other roots.


I believe that a = -3 and b = -6.
I am unsure of how to find the other roots.

Please show how you found the answer.

  1. Bosnian

    For x = 3

    x ^ 4 + a x ^ 3 + a x ^ 2 + 11 x + b = 0

    Put x = 3 in this equation:

    3 ^ 4 + a * 3 ^ 3 + a * 3 ^ 2 + 11 * 3 + b = 0

    81 + 27 a + 9 a + 33 + b = 0

    36 a + b + 114 = 0


    For x = - 2

    x ^ 4 + a x ^ 3 + a x ^ 2 + 11 x + b = 0

    Put x = - 2 in this equation:

    ( - 2 ) ^ 4 + a * ( - 2 ) ^ 3 + a * ( - 2 ) ^ 2 + 11 * ( - 2 ) + b = 0

    16 - 8 a + 4 a - 22 + b = 0

    - 4 a + b - 6 = 0


    Now you must solve system:

    36 a + b + 114 = 0

    - 4 a + b - 6 = 0

    The solutions are: a = - 3 , b = - 6

    So equation x ^ 4 + a x ^ 3 + a x ^ 2 + 11 x + b = 0 becomes:

    x ^ 4 - 3 x ^ 3 - 3 x ^ 2 + 11 x - 6 = 0


    We can take a polynomial of degree 4, such as:

    ( x - p ) ( x - q ) ( x - r ) ( x - s )

    where: p , q , r , s are roots of polynomials

    In this casse you know two roots : p = 3 and q = - 2

    Now ( x - p ) ( x - q ) ( x - r ) ( x - s ) becomes:

    ( x - 3 ) [ x - ( - 2 ) ] ( x - r ) ( x - s ) =

    ( x - 3 ) ( x + 2 ) ( x - r ) ( x - s ) =

    ( x ^ 2 - x - 6 ) ( x - r ) ( x - s ) = 0

    If

    x ^ 4 - 3 x ^ 3 - 3 x ^ 2 + 11 x - 6 = ( x ^ 2 - x - 6 ) ( x - r ) ( x - s )

    then you must do long division:

    ( x ^ 4 - 3 x ^ 3 - 3 x ^ 2 + 11 x - 6 ) / ( x ^ 2 - x - 6 ) = x ^ 2 - 2 x + 1

    This mean ( x - r ) ( x - s ) = x ^ 2 - 2 x + 1

    If ( x - r ) ( x - s ) = 0 then x ^ 2 - 2 x + 1 also = 0

    x ^ 2 - 2 x + 1 = 0

    Solution: x = 1

    Or:

    x ^ 2 - 2 x + 1 = ( x - r ) ( x - s ) = ( x - 1 ) ( x - 1 )

    Factors of your polynomial:

    x ^ 4 - 3 x ^ 3 - 3 x ^ 2 + 11 x - 6 = ( x - 3 ) ( x + 2 ) ( x - 1 ) ( x - 1 )

    All this mean:

    Your polynomialal have 3 roots:

    x = - 2 , x = 3 and a double root x = 1

  2. Steve

    Divide by (x^2-x-6) and you get
    quotient: x^2+(a+1)x+(2a+7)
    remainder: (8a+24)x + b+6(2a+7)

    For the remainder to be zero, we need

    a = -3
    b = -6

    So, x^4-3x^3-3x^2+11x-6
    = (x+2)(x-3)(x^2-2x+1)

    The other roots are 1,1

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