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Math Help - Quadratic Equation Part 3

  1. #1
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    Quadratic Equation Part 3

    1. Express \ln(2\sqrt e)-\frac{1}{3}\ln(\frac{8}{e})-\ln(\frac{e}{3}) in the form c+ln d, where c and d are rational numbers.

    2. Given p(x)=x^4+2x^3+ax^2+bx-60 has factor x+2 and when p(x) is divided by x+3, the remainder is 60. Find a and b.

    2a. Show that x-3 is a factor of p(x) and find q(x) so that p(x)=(x+2)(x-3)q(x)

    2b. Show that q(x)=0 does not possess real roots and find the set of x for which p(x)>0
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  2. #2
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    (1)
     \ln(2\sqrt e)-\frac{1}{3}\ln(\frac{8}{e})-\ln(\frac{e}{3})

    use identity
     \ln A- \ln B=\ln \left ( \frac{A}{B} \right ) \quad , \quad  \ln \ e = 1 \quad and \quad \ln A+ \ln B=\ln (AB)
    see spoiler if stuck
    Spoiler:

     \ln(2\sqrt e)-\frac{1}{3}\ln(\frac{8}{e})-\ln(\frac{e}{3}) =\ln(2 \sqrt e)- \ln(\frac{8}{e})^{\frac{1}{3}}-\ln(\frac{e}{3})
    =\ln(2 e^{ \frac{1}{2}})- \ln( 2 e^{ \frac{-1}{3}})-\ln( \frac{e}{3})= \ln \left ( \frac {2 e^{ \frac{1}{2}}}{2 e^{ \frac{-1}{3}}} \right )-\ln( \frac{e}{3})
     =\ln (e^{\frac{5}{6}})-\ln( \frac{e}{3})=\ln \left (\frac {e^{\frac{5}{6}}}{\frac{e}{3}} \right )=\ln (3 e^{\frac{-1}{6}})
     =\ln (3)+\ln (e^{\frac{-1}{6}})=\ln (3)-\frac{1}{6}\ln (e)=\ln (3)-\frac{1}{6}

    (2)
    Given has factor x+2
     \therefore p(-2) =0 , put=-2 and p(x)=0 and find first equation
    Spoiler:

    0=(-2)^4+2(-2)^3+a(-2)^2+b(-2)-60
     2a-b=30 \quad ....(1)

    and when p(x) is divided by x+3, the remainder is 60.
    \therefore p(-3)=60 , put=-3 and p(x)=60 and find second equation
    Spoiler:

    60=(-3)^4+2(-3)^3+a(-3)^2+b(-3)-60
    3a-b=31 \quad ....(2)

    solve these two equation to find a and b
    i got a= 1, b=-28
    \boxed {\therefore p(x)=x^4+2x^3+x^2-28x-60 }

    continue....
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