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Thread: factor the equation

  1. #1
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    factor the equation

    3q^2 + 16q + 5
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  2. #2
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    I look for a solution of the form (3q-a)(q-b). Expanding $\displaystyle 3q^2 - (3b+a)q +ab$. Equating the coefficients of like terms gives ab=5 and 3b+a=-16. Thinking for a moment will convince you that b=-5 and a=-1 satisfy these equations. So $\displaystyle 3q^2 + 16q + 5= (3q+1)(q+5)$.
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  3. #3
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    No......my previous answer is wrong......sorry about tht!!!

    3q^2 + 16q +5

    3q^2 +15q +q +5

    3q( q+5) +1 (q+5)

    (3q+1) (q+5)
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  4. #4
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    Quote Originally Posted by kenzie103109 View Post
    3q^2 + 16q + 5
    $\displaystyle P(q)=3q^2 + 16q + 5$

    The rational roots theorem suggests that $\displaystyle \pm 5$, $\displaystyle \pm 1/3$ and $\displaystyle \pm 5/3$ are the only possible rational roots for this quadratic in $\displaystyle q$. Descartes rule of signs tells you it has no positive roots, so that reduces the list to $\displaystyle -5, -1/3$ and $\displaystyle -5/3$. Now trying these out shows that $\displaystyle P(-5)=0$ and $\displaystyle P(-1/3)=0$, so we have:

    $\displaystyle P(q)=k(q+1/3)(q+5)$

    and to match the leading coefficient of $\displaystyle P(q)$ we must have $\displaystyle k=3$ so:

    $\displaystyle P(q)=(3q+1)(q+5)$

    CB
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