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Thread: algebra

  1. #1
    Junior Member
    May 2008


    determine the value k so that x-4 is a factor of $\displaystyle x^2+8x+k$
    i know that answer is going o be 12, i got it by dividing x-4 into $\displaystyle x^2+8x+k$ but im not sure about the remainder, bcz i got k+48
    does that change the answer.

    $\displaystyle m^2-k^2+6k-9$ has a factor of (m-k+3) what is the other
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  2. #2
    May 2009
    $\displaystyle x^2+8x+k$ to factor this with (x-4) being a factor we would get (x-4)(x+y)=$\displaystyle x^2+(-4+y)x-4y$ and since the coefficients have to match up, $\displaystyle (-4+y)=8$ and -4y=k

    From (-4+y)=8 we see y=12, so -4(12)=-48=k
    and you can see that $\displaystyle x^2+8x-48=(x-4)(x+12)$

    $\displaystyle m^2-k^2+6k-9=m^2-(k^2-6k+9)=m^2-(k-3)^2$ which is the difference of two perfect squares and $\displaystyle x^2-y^2=(x-y)(x+y)$ so $\displaystyle m^2-(k-3)^2=(m-(k-3))(m+(k-3))=(m-k+3)(m+k-3)$
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