How would you make partial fractions of? #(7x-9)/((x+1)(x-3))# And #(x-11)/((x-4)(x+3))#

1 Answer
Jim G.
Feb 20, 2018

#"see explanation"#

Explanation:

#"since the factors on the denominator are linear we can"#
#"express the rational function as"#

#(7x-9)/((x+1)(x-3))=A/(x+1)+B/(x-3)#

#"multiply both sides by "(x+1)(x-3)" to obtain"#

#rArr7x-9=A(x-3)+B(x+1)#

#"using "color(blue)"Heaviside's cover up rule"#

#"that is use the zeros of the factors and substitute"#
#"into both sides to solve for A and B"#

#x=-1to-16=-4A+B(0)rArrA=4#

#x=3to12=A(0)+4BrArrB=3#

#rArr(7x-9)/((x+1)(x-3))=4/(x+1)+3/(x-3)#

#color(blue)"Similarly"#

#(x-11)/((x-4)(x+3))=A/(x-4)+B/(x+3)#

#"multiply both sides by "(x-4)(x+3)#

#rArrx-11=A(x+3)+B(x-4)#

#x=-3to-14=A(0)-7BrArrB=2#

#x=4to-7=7A+B(0)rArrA=-1#

#rArr(x-11)/((x-4)(x+3))=2/(x+3)-1/(x-4)#

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