#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)#
