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Math Help - Polynomial

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    Polynomial

    The Polynomial x^4 + 5x +a is denoted by p(x). It is Given that x - x +3 is a factor of p(x).

    a) Find the Value of a and factorize completely.

    b) Hence state the number of real roots of the equation p(x) = 0 , Justifying your answer.
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    Quote Originally Posted by Shannon View Post
    The Polynomial x^4 + 5x +a is denoted by p(x). It is Given that x - x +3 is a factor of p(x).

    a) Find the Value of a and factorize completely.

    b) Hence state the number of real roots of the equation p(x) = 0 , Justifying your answer.
    Hi Shannon,

    Using polynomial long division, it can be determined that a = -6.

    Divide x^4+5x+a by x^2-x+3 to determine the value of a than makes the remainder equal to zero.

    The factors of p(x)=x^4+5x-6 are p(x)=(x^2-x+3)(x^2+x-2)\Longrightarrow p(x)=(x^2-x+3)(x+2)(x-1)

    There are two real zeros at -2 and 1.

    There are two imaginary zeros when you solve x^2-x+3=0
    Last edited by masters; February 23rd 2010 at 03:56 AM.
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  3. #3
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    Quote Originally Posted by Shannon View Post
    The Polynomial x^4 + 5x +a is denoted by p(x). It is Given that x - x +3 is a factor of p(x).

    a) Find the Value of a and factorize completely.

    b) Hence state the number of real roots of the equation p(x) = 0 , Justifying your answer.
    Hi Shannon,

    x^4+5x+a=(x^2-x+3)(x^2+bx+c)=x^4+bx^3+cx^2-x^3-bx^2-cx+3x^2+3bx+3c

    x^4+(0)x^3+(0)x^2+5x+a=x^4+(b-1)x^3+(c-b+3)x^2+(3b-c)x+3c

    \Rightarrow\ b=1,\ \Rightarrow\ c=-2,\ \Rightarrow\ a=-6

    x^4+5x-6=(x^2-x+3)(x^2+x-2)=(x^2-x+3)(x+2)(x-1)

    There are 2 real roots, as the roots of x^2-x+3 are

    x=\frac{1\pm\sqrt{1-4(3)}}{2}

    which are complex
    Last edited by Archie Meade; February 22nd 2010 at 02:18 PM. Reason: small typo
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