1) Express (-2x^2-7x-13)/(x^3+4x-3(x^2)-12) in partial fractions.

2) Variables x and y are related by the equation y(e^a)=x^(k-1) where a and k are
constants. When the graph of lnxy and lnx is drawn a straight line is obtained. Given that the line passes through (-2,11) and (1,2), find the value of a and k. (Is a=-5 and k=-3?) Thanks!

Question

1) Express (-2x^2-7x-13)/(x^3+4x-3(x^2)-12) in partial fractions.

2) Variables x and y are related by the equation y(e^a)=x^(k-1) where a and k are
 constants. When the graph of lnxy and lnx is drawn a straight line is obtained. Given that the line passes through (-2,11) and (1,2), find the value of a and k. (Is a=-5 and k=-3?) Thanks!

zepdrix · Answer

$$\Large \frac{-2x^2-7x-13}{x^3-3x^2+4x-12}$$Hmmmmmmmmm were you able to factor it?
I'm trying to remember how to factor a cube -_-

anonymous · Answer

hmmmm

anonymous · Answer

u can use splitting the mid term

anonymous · Answer

or else synthetic division to factor cubivc polynomials

anonymous · Answer

kkk

anonymous · Answer

i kind of tried factorising and manipulating  it and i got 2+((4x^2-15x+11)/(x^3-3x^2+4x-12))

anonymous · Answer

kkk

anonymous · Answer

then group and factorise it

anonymous · Answer

im kinda stuck after this step cause i i do not know how to make it into a partial fraction after this>.<

dumbcow · Answer

factor by grouping
$$(x^{3} -3x^{2})+(4x-12) = x^{2}(x-3) +4(x-3) = (x^{2}+4)(x-3)$$
$$\frac{-2x^{2}-7x-13}{(x^{2}+4)(x-3)} = \frac{Ax +B}{x^{2}+4}+\frac{C}{x-3}$$

dumbcow · Answer

as far as 2nd part of post, you are correct
$$\ln (xy) = k \ln x -a$$
the line going through 2 points is:
y = -3x +5

thus
k = -3
a = -5