Question

Write the partial fraction decomposition of the rational expression. 11x - 48 x2 - 9x + 20

Answer 1

Solution-

We have given the rational expression as-

Now, let us first write denominator in factor form.

$\frac{11x-48}{x^2-5x-4x+20}$

$=\frac{11x-48}{x(x-5)-4(x-5)}$

$=\frac{11x-48}{(x-5)(x-4)}$

Now, this can be written in the form as partial fraction as

$\frac{11x-48}{(x-5)(x-4)}=\frac{A}{x-5}+\frac{B}{x-4}$ .....(1)

Taking lcm on right side.

$\frac{11x-48}{(x-5)(x-4)}=\frac{A(x-4)+B(x-5)}{(x-5)(x-4)}$

Or

Taking Denominator of the sides.

$11x-48=A(x-4)+B(x-5)$

On putting x =4, we get

$11(4)-48=A(4-4)+B(4-5)$

$44-48=A(0)+B(-1)$

$-4=-B$

$4=B$

Or

$B=4$

Again putting x = 5, we get

$11(5)-48=A(5-4)+B(5-5)$

$55-48=A(1)+B(0)$

$7=A$

Or

$A=7$

On putting A=7 and B= 4 (as calculated above) in equation (1), we get

$\frac{11x-48}{(x-5)(x-4)}=\frac{7}{x-5}+\frac{4}{x-4}$

Hence, the given rational expression can be written in partial fraction decomposition form as

$\frac{11x-48}{x^2-9x+20}=\frac{7}{x-5}+\frac{4}{x-4}$