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W 9.4.35 Find the partial fraction decomposition for the following rational expression. 3x® +6x4 - 72x2 + 5x² - 10x - 120 x²+2x² – 24 3x + 6x4 - 72x2 + 5x² - 10x - 120 x3 + 2x² - 24x 1.0.0.0 II X-4 X+6

Answer 1

9.4.35. solution:

We have the given expression as

$\small \frac{3x^{5}+6x^{4}-72x^{3}+5x^{2}-10x-120}{x^{3}+2x^{2}-24x}$

The given expression is an improper fraction since the degree of numerator is greater than the denominator. So, to make it easier, here we use the long division method to find the different parts i.e. whole and fraction parts.

Therefore, the given expression can be written as

$\small 3x^{2}+\frac{5x^{2}-10x-120}{x^{3}+2x^{2}-24x}$

= 3.2 + 5.12 - 10x - 120 (3-4)3 + 6)

Now, we need to find the partial fraction of

$\small \frac{5x^{2}-10x-120}{x(x-4)(x+6)}$

Then, using partial fraction decomposition, we get,

$\small \frac{5x^{2}-10x-120}{x(x-4)(x+6)}=\frac{A}{x}+\frac{B}{x-4}+\frac{C}{x+6}$

$\small \implies 5x^{2}-10x-120=A(x-4)(x+6)+Bx(x+6)+Cx(x-4)$

$\small \implies 5x^{2}-10x-120=Ax^{2}+2Ax-24A+Bx^{2}+6Bx+Cx^{2}-4Cx$

$\small \implies 5x^{2}-10x-120=(A+B+C)x^{2}+(2A+6B-4C)x-24A$

Comparing the coefficients of both sides, we get,

$\small A+B+C=5$

$\small 2A+6B-4C=-10$

$\small -24A=-120\\ \implies A=\frac{120}{24}=5$

now,

$\small 5+B+C=5\\ \implies B+C=5-5=0\\ \implies C=-B$

and,

$\small 2(5)+6B-4(-B)=-10\\ \implies 10+6B+4B=-10\\ \implies 10B=-10-10=20\\ \implies B=\frac{-20}{10}=-2$

also,

$\small C=-(-2)=2$

Hence, the required expression is

$\small =3x^{2}+\frac{5}{x}-\frac{2}{x-4}+\frac{2}{x+6}$

9.4.39. soln.:

We have the given expression as

$\small \frac{8x^{2}-6x-8}{x^{3}+x^{2}-2x}=\frac{8x^{2}-6x-8}{x(x-1)(x+2)}$

Now, using the partial fraction decomposition, we get,

$\small \frac{8x^{2}-6x-8}{x(x-1)(x+2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+2}$

$\small \implies 8x^{2}-6x-8=A(x-1)(x+2)+Bx(x+2)+Cx(x-1)$

$\small \implies 8x^{2}-6x-8=Ax^{2}+Ax-2A+Bx^{2}+2Bx+Cx^{2}-Cx$

$\small \implies 8x^{2}-6x-8=Ax^{2}+Bx^{2}+Cx^{2}+Ax+2Bx-Cx-2A$

$\small \implies 8x^{2}-6x-8=\left (A+B+C \right )x^{2}+(A+2B-C)x-2A$

Comparing the coefficients of both sides, we get,

$\small A+B+C=8$

$\small A+2B-C=-6$

$\small -2A=-8\\ \implies A=\frac{-8}{-2}=4$

and,

$\small 4+B+C=8\\ \implies B+C=8-4=4\\ \implies C=4-B$

also,

$\small 4+2B-(4-B)=-6\\ \implies 2B-4+B=-6-4=-10\\ \implies 3B=-10+4=-6\\ \implies B=\frac{-6}{3}=-2$

again,

$\small C=4-(-2)=4+2=6$

Hence, the required expression is

$\small \frac{4}{x}-\frac{2}{x-1}+\frac{6}{x+2}$