Binomial expansion with partial fractions In a nutshell Partial fractions can be used to break apart more complicated expressions into sums of simpler expressions, and hence help you find their expansions.
Finding the expansion of more complicated expressions Given an expression of the form P ( x ) Q ( x ) \dfrac{P(x)}{Q(x)} Q ( x ) P ( x ) P ( x ) P(x) P ( x ) Q ( x ) Q(x) Q ( x ) A ( a x + b ) \dfrac{A}{(ax + b)} ( a x + b ) A A ( a x + b ) 2 \dfrac{A}{(ax + b)^2} ( a x + b ) 2 A ∣ x ∣ < ∣ a b ∣ |x| < \left|\dfrac{a}{b}\right| ∣ x ∣ < b a n n n
Example 1 Find the first three terms in the expansion of 7 x 2 − 17 x − 29 ( 2 − x ) 2 ( 3 + 2 x ) \dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)} ( 2 − x ) 2 ( 3 + 2 x ) 7 x 2 − 17 x − 29 x x x
Use partial fractions:
7 x 2 − 17 x − 29 ( 2 − x ) 2 ( 3 + 2 x ) = A 2 − x + B ( 2 − x ) 2 + C 3 + 2 x 7 x 2 − 17 x − 29 = ( 3 + 2 x ) ( 2 − x ) A + ( 3 + 2 x ) B + ( 2 − x ) 2 C 7 x 2 − 17 x − 29 = ( C − 2 A ) x 2 + ( A + 2 B − 4 C ) x + ( 6 A + 3 B + 4 C ) \begin{aligned}\dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)} &= \dfrac{A}{2-x} + \dfrac{B}{(2-x)^2} + \dfrac{C}{3 + 2x}\\\\7x^2 - 17x - 29&=(3+2x)(2-x)A + (3+2x)B + (2-x)^2C\\\\7x^2 - 17x - 29 &= (C - 2A)x^2 + (A + 2B -4C)x + (6A + 3B + 4C)\end{aligned} ( 2 − x ) 2 ( 3 + 2 x ) 7 x 2 − 17 x − 29 7 x 2 − 17 x − 29 7 x 2 − 17 x − 29 = 2 − x A + ( 2 − x ) 2 B + 3 + 2 x C = ( 3 + 2 x ) ( 2 − x ) A + ( 3 + 2 x ) B + ( 2 − x ) 2 C = ( C − 2 A ) x 2 + ( A + 2 B − 4 C ) x + ( 6 A + 3 B + 4 C )
Compare coefficients to obtain the system of linear equations:
7 = C − 2 A − 17 = A + 2 B − 4 C − 29 = 6 A + 3 B + 4 C \begin{aligned}7 &= C - 2A\\-17 &= A + 2B - 4C\\-29 &= 6A + 3B + 4C\end{aligned} 7 − 17 − 29 = C − 2 A = A + 2 B − 4 C = 6 A + 3 B + 4 C
Solve these to get the values:
A = − 3 B = − 5 C = 1 \begin{aligned}A &= -3\\B &= -5\\C &= 1\end{aligned} A B C = − 3 = − 5 = 1
So, you have that:
7 x 2 − 17 x − 29 ( 2 − x ) 2 ( 3 + 2 x ) = − 3 2 − x + − 5 ( 2 − x ) 2 + 1 3 + 2 x \dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)} = \dfrac{-3}{2-x} + \dfrac{-5}{(2-x)^2} + \dfrac{1}{3 + 2x} ( 2 − x ) 2 ( 3 + 2 x ) 7 x 2 − 17 x − 29 = 2 − x − 3 + ( 2 − x ) 2 − 5 + 3 + 2 x 1
The expansion of − 3 2 − x = − 3 2 ( 1 − x 2 ) − 1 \dfrac{-3}{2-x} = -\dfrac32\left(1 - \dfrac{x}{2}\right)^{-1} 2 − x − 3 = − 2 3 ( 1 − 2 x ) − 1 is:
− 3 2 ( 1 − x 2 ) − 1 = − 3 2 [ 1 − ( − x 2 ) + ( − 1 ) ( − 2 ) 2 ! ( − x 2 ) 2 + … ] = − 3 2 − 3 x 4 − 3 x 2 8 + … \begin{aligned}-\dfrac32\left(1 - \dfrac{x}{2}\right)^{-1} &= -\dfrac32\left[1 - \left(-\dfrac{x}{2}\right) + \dfrac{(-1)(-2)}{2!}\left(-\dfrac{x}{2}\right)^2 + \dots\right]\\&= -\dfrac32 - \dfrac{3x}{4} - \dfrac{3x^2}{8} + \dots\end{aligned} − 2 3 ( 1 − 2 x ) − 1 = − 2 3 [ 1 − ( − 2 x ) + 2 ! ( − 1 ) ( − 2 ) ( − 2 x ) 2 + … ] = − 2 3 − 4 3 x − 8 3 x 2 + …
And this is valid while ∣ x ∣ < 2 |x| < 2 ∣ x ∣ < 2
The expansion of − 5 ( 2 − x ) 2 = − 5 4 ( 1 − x 2 ) − 2 \dfrac{-5}{(2-x)^2} = -\dfrac54\left(1-\dfrac{x}{2}\right)^{-2} ( 2 − x ) 2 − 5 = − 4 5 ( 1 − 2 x ) − 2
− 5 4 ( 1 − x 2 ) − 2 = − 5 4 [ 1 − 2 ( − x 2 ) + ( − 2 ) ( − 3 ) 2 ! ( − x 2 ) 2 + … ] = − 5 4 − 5 x 4 − 15 x 2 16 + … \begin{aligned}-\dfrac54\left(1 - \dfrac{x}{2}\right)^{-2} &= -\dfrac54\left[1 -2\left(-\dfrac{x}{2}\right) + \dfrac{(-2)(-3)}{2!}\left(-\dfrac{x}{2}\right)^2 + \dots\right]\\&= -\dfrac54 - \dfrac{5x}{4} - \dfrac{15x^2}{16} + \dots\end{aligned} − 4 5 ( 1 − 2 x ) − 2 = − 4 5 [ 1 − 2 ( − 2 x ) + 2 ! ( − 2 ) ( − 3 ) ( − 2 x ) 2 + … ] = − 4 5 − 4 5 x − 16 15 x 2 + …
And this is valid while ∣ x ∣ < 2 |x| < 2 ∣ x ∣ < 2
The expansion of 1 3 + 2 x = 1 3 ( 1 + 2 x 3 ) − 1 \dfrac{1}{3 + 2x} = \dfrac13\left(1 + \dfrac{2x}{3}\right)^{-1} 3 + 2 x 1 = 3 1 ( 1 + 3 2 x ) − 1
1 3 ( 1 + 2 x 3 ) − 1 = 1 3 [ 1 − ( 2 x 3 ) + ( − 1 ) ( − 2 ) 2 ! ( 2 x 3 ) 2 + … ] = 1 3 − 2 x 9 + 4 x 2 27 + … \begin{aligned}\dfrac13\left(1 + \dfrac{2x}{3}\right)^{-1} &= \dfrac13\left[1 - \left(\dfrac{2x}{3}\right) + \dfrac{(-1)(-2)}{2!}\left(\dfrac{2x}{3}\right)^2 + \dots\right]\\&= \dfrac13 - \dfrac{2x}{9} + \dfrac{4x^2}{27} + \dots\end{aligned} 3 1 ( 1 + 3 2 x ) − 1 = 3 1 [ 1 − ( 3 2 x ) + 2 ! ( − 1 ) ( − 2 ) ( 3 2 x ) 2 + … ] = 3 1 − 9 2 x + 27 4 x 2 + …
And this is valid while ∣ x ∣ < 3 2 |x| < \dfrac32 ∣ x ∣ < 2 3
Adding these together, you have that the first three terms in the expansion of 7 x 2 − 17 x − 29 ( 2 − x ) 2 ( 3 + 2 x ) \dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)} ( 2 − x ) 2 ( 3 + 2 x ) 7 x 2 − 17 x − 29
− 29 12 − 20 x 9 − 503 x 2 432 + … -\dfrac{29}{12} - \dfrac{20x}{9} - \dfrac{503x^2}{432} + \dots − 12 29 − 9 20 x − 432 503 x 2 + …
And this is only valid when all three other series have valid expansions, i.e. while ∣ x ∣ < 3 2 |x| < \dfrac32 ∣ x ∣ < 2 3
Therefore, the first three terms in the expansion of 7 x 2 − 17 x − 29 ( 2 − x ) 2 ( 3 + 2 x ) \dfrac{7x^2 - 17x - 29}{(2-x)^2(3+2x)} ( 2 − x ) 2 ( 3 + 2 x ) 7 x 2 − 17 x − 29 − 29 12 − 20 x 9 − 503 x 2 432 + … ‾ \underline{-\dfrac{29}{12} - \dfrac{20x}{9} - \dfrac{503x^2}{432} + \dots} − 12 29 − 9 20 x − 432 503 x 2 + … ∣ x ∣ < 3 2 ‾ \underline{|x| < \dfrac32} ∣ x ∣ < 2 3
