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Thread: polynomials

  1. #1
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    polynomials

    10x^2-3x-3= A(x^2+1)+(Bx+C) (x-1)

    find A B C


    y=kx-5
    y=6x-x^2-6

    solve for k
    Last edited by amath9; Jun 12th 2009 at 08:05 PM.
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  2. #2
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    Quote Originally Posted by amath9 View Post
    10x^2-3x-3= A(x^2+1)+(Bx+C) (x-1)

    find A B C


    y=kx-5
    y=6x-x^2-6

    solve for k


    $\displaystyle 10x^2-3x-3 =A(x^2+1)+(Bx+C)(x-1) $

    $\displaystyle 10x^2-3x-3=Ax^2+A +Bx^2-Bx+xC-C$

    the factors which multiply with $\displaystyle x^2$ in the right side should equal to the factor of $\displaystyle x^2$ in the left side so you will have

    $\displaystyle 10=A+B$

    and the factors which are multiply with x in the right side should equal to the factor with x in the left side so you will have

    $\displaystyle -3=-B+C$

    and finally the terms with no x in the right side should equal to the terms with no x in the left side so you will have this

    $\displaystyle -3 = A -C$

    now you have three equation with three variables so you can solve them

    $\displaystyle 10=A+B$

    $\displaystyle -3=-B+C$

    $\displaystyle -3 = A -C$

    find the sum of the second and the third

    $\displaystyle -6=A-B$ find the sum of this and the first one

    $\displaystyle 4=2A \Rightarrow A=2 $ the rest for you
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  3. #3
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    Quote Originally Posted by amath9 View Post
    10x^2-3x-3= A(x^2+1)+(Bx+C) (x-1)

    find A B C
    Multiply out the right side: $\displaystyle 10x^2- 3x- 3= Ax^2+ A+ Bx^2- Bx+ Cx- C$.
    $\displaystyle 10x^2- 3x- 3= (A+ B)x^2+ (C- B)x- C$
    In order that these be equal for all x, corresponding coefficients must be equal: 10= A+ B, -3= C- B, and -3= -C. Solve for A, B, and C.


    y=kx-5
    y=6x-x^2-6

    solve for k
    $\displaystyle y= kx- 5= 6x- x^2- 6$. Because the left side does not have an "$\displaystyle x^2$" term, there is NO value of k that will make this true for all x.
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