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Math Help - Finding Values of A and B

  1. #1
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    Finding Values of A and B

    Determine a and b so that (x-1)^2 divides evenly into the polynomial
    f(x)=ax^4+bx^3+1.
    First you must divide x-1 into f(x), the divide x-1 again into quotient.
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  2. #2
    Newbie I4talent's Avatar
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    Quote Originally Posted by KingV15 View Post
    Determine a and b so that (x-1)^2 divides evenly into the polynomial
    f(x)=ax^4+bx^3+1.
    First you must divide x-1 into f(x), the divide x-1 again into quotient.
    Let ax^4+bx^3+1 = (x-1)(x-1)(cx^2+dx+e)<br />

    <br />
ax^4+bx^3+1 = (x-1)(x-1)(cx^2+dx+e)

    ax^4+bx^3+1 = (x^2-2x+1)(cx^2+dx+e)

    ax^4+bx^3+1 = x^2(cx^2+dx+e)-2x(cx^2+dx+e)+1(cx^2+dx+e)

    ax^4+bx^3+1 = cx^4+dx^3+ex^2-2cx^3-2dx^2-2ex+cx^2+dx+e

    ax^4+bx^3+1 = cx^4+dx^3-2cx^3+ex^2-2dx^2cx^2-2ex+dx+e

    ax^4+bx^3+1 = (c)x^4+(d-2c)x^3+(e-2d+c)x^2+(-2e+d)x+e

    Comparing the coefficients:

    \boxed{e = 1}.

    -2e+d = 0 \Rightarrow -2(1)+d = 0 \Rightarrow -2+d = 0 \Rightarrow \boxed{d = 2}

    e-2d+c = 0 \Rightarrow 1-2(2)+c = 0 \Rightarrow -3+c = 0 \Rightarrow \boxed{c = 3}

    d-2c = b \Rightarrow 2-2(3) = b \Rightarrow 2-6 = b \Rightarrow \boxed{-4 = b}

    a = c \Rightarrow \boxed{a = 3}

    Hence \boxed{a = 3} and \boxed{b = -4}
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