1. First, we factorise the denominator, but since this is done, we skip to the next step: Writing a partial fraction for each factor:
(-7x^2-15x-8)/((x+3)(x^2+4))=a/(x+3) + (bx+c)/(x^2+4)
2. Multiply out the factors to remove the fractions:
-7x^2-15x-8=a(x^2+4) + (bx+c)(x+3)
3. Solve the equation for the constants:
-7x^2-15x-8= (-7x-8)(x+1)
(-7x-8)(x+1)= a(x^2+4) + (bx+c)(x+3)
"a") If we set x=-3, then we can solve for a because bx+c=0
(-7*(-3)-8)(-3+1) = a((-3)^2+4)
-26=13a
a=-2
(-7x-8)(x+1) = -2(x^2+4) + (bx+c)(x+3)
"b") If we set x=0, we can solve for c because bx=0
(-8)(+1) = -2(+4)+c(3)
-8=-8+3c
3c=0, c=0
(-7x-8)(x+1) = -2(x^2+4) + (bx)(x+3)
"c") If we set x=1, we can now solve for b relatively simply
(-7-8)(1+1) = -2(1+4) +b(1+3)
-30 = -10 + 4b
-20=4b
b=-5
4.Now we substitute the values back into the partial fractions:
(-7x^2-15x-8)/((x+3)(x^2+4))=(-2)/(x+3) + (-5x)/(x^2+4)
