• Oct 27th 2009, 03:05 AM
cloud5
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
• Oct 27th 2009, 05:02 AM
ramiee2010
(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 http://www.mathhelpforum.com/math-he...4582042a-1.gif 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....