1. First, we factorise the denominator, but since this is done, we skip to the next step: Writing a partial fraction for each factor:
−7x2−15x−8(x+3)(x2+4)=ax+3+bx+cx2+4
2. Multiply out the factors to remove the fractions:
−7x2−15x−8=a(x2+4)+(bx+c)(x+3)
3. Solve the equation for the constants:
−7x2−15x−8=(−7x−8)(x+1)
(−7x−8)(x+1)=a(x2+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(x2+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(x2+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:
−7x2−15x−8(x+3)(x2+4)=−2x+3+−5xx2+4
