Express the following improper fraction as partial fraction. 22 - 17 x + 21 x 2 - 4 x 3 x - 4 x 2 + 1

We have 22 - 17 x + 21 x 2 - 4 x 3 x - 4 x 2 + 1 = ( 22 - 17 x + 21 x 2 - 4 x 3 ) ( x 3 + x - 4 x 2 - 4 ) By long Division method, x 3 − 4 x 2 + x − 4 − − 4 + 21 x 2 − 17 x + 22 x 3 − 4 x 2 + x − 4 − 4 x 3 + 21 x 2 − 17 x + 22 x 3 − 4 x 2 + x − 4 − − 4 x 3 + 21 x 2 − 17 x + 22 − 4 x 3 + 16 x 2 − 4 x + 16 5 x 2 − 13 x + 6 So, - 4 x 3 + 21 x 2 - 17 x + 22 ( x - 4 ) x 2 + 1 = - 4 + 5 x 2 - 13 x + 6 ( x - 4 ) x 2 + 1 We can write 5 x 2 - 13 x + 6 ( x - 4 ) x 2 + 1 in the partial fraction form as 5 x 2 - 13 x + 6 ( x - 4 ) x 2 + 1 = A x + B x 2 + 1 + C x - 4 where A , B , C ∈ ℝ ⇒ 5 x 2 - 13 x + 6 ( x - 4 ) x 2 + 1 = ( x - 4 ) ( A x + B ) + x 2 + 1 C ( x - 4 ) x 2 + 1 ⇒ 5 x 2 - 13 x + 6 = x 2 A + x 2 C - 4 x A + x B - 4 B + C ⇒ 5 x 2 - 13 x + 6 = x 2 ( A + C ) + x ( - 4 A + B ) - 4 B + C Equating the coefficients of x 2 , x and constant on the sides, we get A + C = 5 . . . ( 1 ) - 4 A + B = - 13 . . . ( 2 ) - 4 B + C = 6 . . . ( 3 ) Solving above equations, we get A = 3 , B = - 1 , C = 2 5 x 2 - 13 x + 6 ( x - 4 ) x 2 + 1 = 3 x - 1 x 2 + 1 + 2 x - 4 And hence, we can write the given improper fraction in the partial fraction as follows - 4 x 3 + 21 x 2 - 17 x + 22 ( x - 4 ) x 2 + 1 = - 4 + 3 x - 1 x 2 + 1 + 2 x - 4

Express the following improper fractions as partial fractions. 4 x 3 + x 2 - 16 x + 7 2 x 3 - 4 x 2 + 2 x

The given improper fraction is 4 x 3 + x 2 - 16 x + 7 2 x 3 - 4 x 2 + 2 x We divide 4 x 3 + x 2 - 16 x + 7 by 2 x 3 - 4 x 2 + 2 x . 2 x 3 − 4 x 2 + 2 x − 2 2 x 3 − 4 x 2 + 2 x − 4 x 3 + x 2 − 16 x + 7 2 x 3 − 4 x 2 + 2 x − 4 x 3 + x 2 − 16 x + 7 4 x 3 − 8 x 2 + 4 x 9 x 2 − 20 x + 7 9 x 2 − 20 x + 7 So, 4 x 3 + x 2 - 16 x + 7 2 x ( x - 1 ) 2 = 2 + 9 x 2 - 20 x + 7 2 x ( x - 1 ) 2 Now, let 9 x 2 2 - 10 x + 7 2 x ( x - 1 ) 2 = A x + B x - 1 + C ( x - 1 ) 2 ⇒ 9 x 2 2 - 10 x + 7 2 x ( x - 1 ) 2 = x ( x - 1 ) B + x C + ( x - 1 ) 2 A x ( x - 1 ) 2 ⇒ 9 x 2 2 - 10 x + 7 2 = x ( x - 1 ) B + x C + ( x - 1 ) 2 A ⇒ 9 x 2 2 - 10 x + 7 2 = x 2 A + x 2 B - 2 x A - x B + x C + A ⇒ 9 x 2 2 - 10 x + 7 2 = x 2 ( A + B ) + x ( - 2 A - B + C ) + A Equating the coefficients of x 2 , x and the constant on both the sides of above equation, we get A + B = 9 2 . . . ( 1 ) - 2 A - B + C = - 10 . . . ( 2 ) A = 7 2 . . . ( 3 ) Solving above three equations simultaneously, we get A = 7 2 , B = 1 , C = - 2 Therefore, 9 x 2 2 - 10 x + 7 2 x ( x - 1 ) 2 = 7 2 x + 1 x - 1 + - 2 ( x - 1 ) 2 Thus, we can express the given improper fraction in terms of partial fraction as 4 x 3 + x 2 - 16 x + 7 2 x 3 - 4 x 2 + 2 x = 2 + 7 2 x + 1 x - 1 + - 2 ( x - 1 ) 2 .

Find the values of A , B , C and D such that 4 x 3 - 9 x 2 + 11 x - 4 x 2 2 x - 1 ≡ A + B x + C x 2 + D 2 x - 1 .

Given that: 4 x 3 - 9 x 2 + 11 x - 4 x 2 2 x - 1 ≡ A + B x + C x 2 + D 2 x - 1 . We have to find the values of A , B , C and D We have, 4 x 3 - 9 x 2 + 11 x - 4 x 2 2 x - 1 = 4 x 3 - 9 x 2 + 11 x - 4 2 x 3 - x 2 4 x 3 - 9 x 2 + 11 x - 4 2 x 3 - x 2 can be written as 4 x 3 - 2 x 2 - 7 x 2 + 11 x - 4 2 x 3 - x 2 = 2 + - 7 x 2 + 11 x - 4 2 x 3 - x 2 So, 4 x 3 - 9 x 2 + 11 x - 4 x 2 2 x - 1 = 2 + - 7 x 2 + 11 x - 4 x 2 2 x - 1 Now consider - 7 x 2 + 11 x - 4 x 2 2 x - 1 = B x + C x 2 + D 2 x - 1 ⇒ - 7 x 2 + 11 x - 4 x 2 2 x - 1 = ( 2 x - 1 ) ( B x + C ) + x 2 D x 2 2 x - 1 ⇒ - 7 x 2 + 11 x - 4 = x 2 2 B + x 2 D - x B + x 2 C - C ⇒ - 7 x 2 + 11 x - 4 = x 2 ( 2 B + D ) + x ( - B + 2 C ) - C Equating the coefficients of x 2 , x and constant on the sides, we get 2 B + D = - 7 . . . ( 1 ) - B + 2 C = 11 . . . ( 2 ) - C = - 4 . . . ( 3 ) Solving above equations, we get B = - 3 , C = 4 , D = - 1 - 7 x 2 + 11 x - 4 x 2 2 x - 1 = - 3 x + 4 x 2 + - 1 2 x - 1 And hence, we can write the given improper fraction in the partial fraction as follows 4 x 3 - 9 x 2 + 11 x - 4 x 2 2 x - 1 = 2 + - 3 x + 4 x 2 + - 1 2 x - 1 Or, 4 x 3 - 9 x 2 + 11 x - 4 x 2 2 x - 1 = 2 + - 3 x + 4 x 2 + - 1 2 x - 1 So, the required values of A , B , C and D are 2 , - 3 , 4 and - 1 .

Express x 2 - 13 x - 5 2 x 3 - 3 x 2 - 3 x + 2 in partial fractions.

We have to express x 2 - 13 x - 5 2 x 3 - 3 x 2 - 3 x + 2 in partial fractions. Let p ( x ) = 2 x 3 - 3 x 2 - 3 x + 2 p ( − 1 ) = 0 Therefore, ( x + 1 ) is one of the factor. The other factor is 2 x 2 − 5 x + 2 ⇒ 2 x 2 − x − 4 x + 2 ⇒ x ( 2 x − 1 ) − 2 ( 2 x − 1 ) ⇒ ( 2 x − 1 ) ( x − 2 ) ⇒ 2 x 3 - 3 x 2 - 3 x + 2 = ( x + 1 ) ( 2 x − 1 ) ( x − 2 ) . So, x 2 - 13 x - 5 2 x 3 - 3 x 2 - 3 x + 2 = x 2 - 13 x - 5 ( x + 1 ) ( 2 x − 1 ) ( x − 2 ) Now consider x 2 - 13 x - 5 ( x + 1 ) ( 2 x − 1 ) ( x − 2 ) = A ( x + 1 ) + B ( 2 x − 1 ) + C ( x − 2 ) ⇒ x 2 - 13 x - 5 ( x + 1 ) ( 2 x − 1 ) ( x − 2 ) = A ( 2 x − 1 ) ( x − 2 ) + B ( x + 1 ) ( x − 2 ) + C ( x + 1 ) ( 2 x − 1 ) ( x + 1 ) ( 2 x − 1 ) ( x − 2 ) ⇒ x 2 - 13 x - 5 = A ( 2 x − 1 ) ( x − 2 ) + B ( x + 1 ) ( x − 2 ) + C ( x + 1 ) ( 2 x − 1 ) ⇒ x 2 - 13 x - 5 = x 2 ( 2 A + B + 2 C ) + x ( - 5 A - B + C ) + 2 A - 2 B - C Equating the coefficients of x 2 , x and constant on the sides, we get 2 A + B + 2 C = 1 . . . ( 1 ) - 5 A - B + C = - 13 . . . ( 2 ) 2 A - 2 B - C = - 5 . . . ( 3 ) Solving above equations, we get A = 1 , B = 5 , C = - 3 So, x 2 - 13 x - 5 ( x + 1 ) ( 2 x − 1 ) ( x − 2 ) = 1 ( x + 1 ) + 5 ( 2 x − 1 ) + - 3 ( x − 2 ) .

If 2 x 3 - 11 x 2 + 12 x + 9 = 2 x + 1 x - 3 2 , then express 24 - x 2 x 3 - 11 x 2 + 12 x + 9 in partial fractions.

Given that: If 2 x 3 - 11 x 2 + 12 x + 9 = 2 x + 1 x - 3 2 We have to express 24 - x 2 x 3 - 11 x 2 + 12 x + 9 in partial fractions. So, 24 - x 2 x 3 - 11 x 2 + 12 x + 9 = 24 - x 2 x + 1 x - 3 2 Now consider 24 - x 2 x + 1 x - 3 2 = A 2 x + 1 + B x - 3 + C x - 3 2 ⇒ 24 - x 2 x + 1 x - 3 2 = A x - 3 2 + B 2 x + 1 x - 3 + 2 x + 1 C 2 x + 1 x - 3 2 ⇒ 24 - x = x 2 ( A + 2 B ) + x ( - 6 A - 5 B + 2 C ) + 9 A - 3 B + C Equating the coefficients of x 2 , x and constant on the sides, we get A + 2 B = 0 . . . ( 1 ) - 6 A - 5 B + 2 C = - 1 . . . ( 2 ) 9 A - 3 B + C = 24 . . . ( 3 ) Solving above equations, we get A = 2 , B = - 1 , C = 3 So, 24 - x 2 x + 1 x - 3 2 = 2 2 x + 1 - 1 x - 3 + 3 x - 3 2 .