the expression in the denominator can be factorised to give the denominator of each partial fraction.
#x^2+5x+4 = (x+4)(x+1)#
the denominators of the partial fractions are #(x+4)# and #(x+1)#.
the numerators are unknown; therefore, it is easiest to label them as two variables (e.g. #a# and #b#).
if this is done, then the partial fractions are #(a)/(x+4) and (b)/(x+1)#.
#(a)/(x+4) + (b)/(x+1) = (3x+18)/(x^2+5x+4)#
to get rid of the denominator of all of the fractions, you can multiply them by #(x^2+5x+4)#, or #(x+4)(x+1)#.
#(a)/(x+4) * (x^2+5x+4) = a(x+1)#
#(b)/(x+1) * (x^2+5x+4) = b(x+4)#
#(3x+18)/(x^2+5x+4) * (x^2+5x+4) = 3x+18#
since #(a)/(x+4) + (b)/(x+1) = (3x+18)/(x^2+5x+4)#,
#a(x+1) + b(x+4) = 3x+18#.
multiplying brackets out gives
#ax + a + bx + 4b = 3x + 18#
then you can group the #x# terms and the constant terms:
#ax + bx + a + 4b = 3x + 18#
#(a+b)x + a + 4b = 3x + 18#
for both sides to be equal, the coefficients of #x# and the coefficients of #1# must be equal to each other.
#(a+b)x = 3x#
#a + 4b = 18#
#a + b = 3#
#a + 4b = 18#
b can be solved for:
#a + 4b = a + b + 15#
#4b = b + 15#
#3b = 15#
#b = 5#
and so can a:
#a + b = 3#
#a + 5 = 3#
#a = 3 - 5#
#a = -2#
these values for #a# and #b# can be substituted into the numerators of the partial fractions #(a)/(x+4)# and #(b)/(x+1)#,
giving #-2/(x+4) and 5/(x+1)#
to check:
#-2/(x+4) + 5/(x+1) = (-2(x+1))/(x^2+5x+4) + (5(x+4))/(x^2+5x+4)#
#= ((-2x - 2) + (5x + 20)) / (x^2+5x+4)#
#= (3x+18)/(x^2+5x+4)#
