32 Partial fraction decomposition; Question 32.9

[Eigenvalue]

32.9. Find the partial fraction decomposition of [tex]\frac{2x+22}{x²+x-12}[/tex]

[Eigenvalue]

a) There is no need for polynomial long division as the degree of the numerator is less than that of the denominator

b) Rewrite [tex]\frac{2x+22}{x²+x-12}[/tex] as [tex]\frac{A}{x+4}[/tex]+[tex]\frac{B}{x-3}[/tex]

[tex]\frac{2x+22}{x²+x-12}[/tex]=[tex]\frac{A(x-3)+B(x+4)}{(x+4)(x-3)}[/tex]

c) As the denominators of the fractions are equal, and the fractions are equal, the numerators of the fraction are equal
2x+22=A(x-3)+B(x+4)

d) Solve for A and B
2x+22=Ax+Bx-3A+4B
A+B=2
4B-3A=22
A=4
B=-2

e)[tex]\frac{2x+22}{x²+x-12}[/tex]=[tex]\frac{4}{x-3}[/tex]+[tex]\frac{3}{x+4}[/tex]