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Thread: Factor/Remainder theorem

  1. #1
    Newbie s2951's Avatar
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    Factor/Remainder theorem

    (x-3) is the Factor of f(x) 4x^2-7x-15
    Help me is this correct of finding the other factor?
    f(x)=l(x)q(x)+R
    f(x)=(x-3)(4x+ax+5)+0
    f(x)=x(4x+ax+5)-3(4x+ax+5)
    f(x)=4x^2+ax^2+5x-12x-3ax-15
    f(x)=(ax+5)=0 (12-3ax)=0
    ax=-5 ax=4
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  2. #2
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    Since $\displaystyle (x-3)$ is a factor,
    so $\displaystyle f(x)=(x-3)(Ax+B)$
    Thus $\displaystyle 4x^{2}-7x-15=(x-3)(Ax+B)$
    By comparison, $\displaystyle A=4,B=-5$
    Hence, $\displaystyle f(x)=(x-3)(4x+5)$

    Anyway, you can get the two factors by the method Cross Multiplication for the quadratic expression, that will be faster.
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  3. #3
    Newbie s2951's Avatar
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    Quote Originally Posted by acc100jt View Post
    Since $\displaystyle (x-3)$ is a factor,
    so $\displaystyle f(x)=(x-3)(Ax+B)$
    Thus $\displaystyle 4x^{2}-7x-15=(x-3)(Ax+B)$
    By comparison, $\displaystyle A=4,B=-5$
    Hence, $\displaystyle f(x)=(x-3)(4x+5)$

    Anyway, you can get the two factors by the method Cross Multiplication for the quadratic expression, that will be faster.
    Ok, then how you get (4x+5) ?
    Last edited by s2951; Dec 2nd 2009 at 12:19 AM.
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  4. #4
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    Hello s2951
    Quote Originally Posted by s2951 View Post
    Ok, then how you get (4x+5) ?
    Quote Originally Posted by s2951 View Post
    (x-3) is the Factor of f(x) 4x^2-7x-15
    Help me is this correct of finding the other factor?
    f(x)=l(x)q(x)+R
    f(x)=(x-3)(4x+ax+5)+0
    f(x)=x(4x+ax+5)-3(4x+ax+5)
    f(x)=4x^2+ax^2+5x-12x-3ax-15
    f(x)=(ax+5)=0 (12-3ax)=0
    ax=-5 ax=4
    I'm not sure where you get the term $\displaystyle ax$ in the line that I've indicated. When you divide $\displaystyle 4x^2-7x-15$ by $\displaystyle x-3$, the result is simply $\displaystyle 4x+5$.

    If I may expand on acc100jt's solution a little:

    If $\displaystyle 4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get:
    $\displaystyle 4x^2-7x-15=Ax^2+(B-3A)x -3B$
    So, when we compare coefficients:
    the coefficient of $\displaystyle x^2:4 = A$
    and
    the constant term: $\displaystyle -15 = -3B \Rightarrow B = 5$
    Check with the coefficient of $\displaystyle x$:
    $\displaystyle B-3A = 5 -12 = -7$, which is correct
    So the other factor is $\displaystyle Ax+B = 4x+5$

    Grandad
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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello s2951

    I'm not sure where you get the term $\displaystyle ax$ in the line that I've indicated. When you divide $\displaystyle 4x^2-7x-15$ by $\displaystyle x-3$, the result is simply $\displaystyle 4x+5$.

    If I may expand on acc100jt's solution a little:

    If $\displaystyle 4x^2-7x-15=(x-3)(Ax+B)$, then when we multiply out the RHS we get:
    $\displaystyle 4x^2-7x-15=Ax^2+(B-3A)x -3B$
    So, when we compare coefficients:
    the coefficient of $\displaystyle x^2:4 = A$
    and
    the constant term: $\displaystyle -15 = -3B \Rightarrow B = 5$
    Check with the coefficient of $\displaystyle x$:
    $\displaystyle B-3A = 5 -12 = -7$, which is correct
    So the other factor is $\displaystyle Ax+B = 4x+5$

    Grandad
    Thanks, just solved and its correct. The letter "A" is just symbol to find the value, its redundant.
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  6. #6
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    Smile

    Here's another way,since $\displaystyle (x-3)$ is a factor,$\displaystyle -3$ is a root.
    We can do the following,
    $\displaystyle f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\times3+15$$\displaystyle =4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$
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  7. #7
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    Hello Raoh
    Quote Originally Posted by Raoh View Post
    Here's another way,since $\displaystyle (x-3)$ is a factor,$\displaystyle -3$ is a root.
    We can do the following,
    $\displaystyle f(x)-f(-3)= 4x^2-7x-15-4(-3)^{2}+7\times3+15$$\displaystyle =4(x^2-(-3)^2)-7(x-1)=4(x+3)(x-3)-7(x-3)=(x-3)(4x+5)$
    I think what you mean is:
    Since $\displaystyle (x-3)$ is a factor, $\displaystyle +3$ is a root $\displaystyle \Rightarrow f(3)=0$
    So:
    $\displaystyle f(x) = f(x) - f(3) = 4(x^2-3^2) - 7(x-3) -(15-15)$
    $\displaystyle =4(x+3)(x-3) -7(x-3)$

    $\displaystyle =(4x+12)(x-3) -7(x-3)$

    $\displaystyle =(4x+12-7)(x-3)$

    $\displaystyle =(4x+5)(x-3)$
    Grandad
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  8. #8
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    Quote Originally Posted by Grandad View Post
    Hello RaohI think what you mean is:
    Since $\displaystyle (x-3)$ is a factor, $\displaystyle +3$ is a root $\displaystyle \Rightarrow f(3)=0$
    So:
    $\displaystyle f(x) = f(x) - f(3) = 4(x^2-3^2) - 7(x-3) -(15-15)$
    $\displaystyle =4(x+3)(x-3) -7(x-3)$

    $\displaystyle =(4x+12)(x-3) -7(x-3)$

    $\displaystyle =(4x+12-7)(x-3)$

    $\displaystyle =(4x+5)(x-3)$
    Grandad
    indeed,my mistake.
    thanks "Grandad".
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