Question

resolved the following into partial fractions: 5x - 2/ x2 - 2x - 8​

Answer 1

Answer:

3/x-4+ 2/x+2

Step-by-step explanation:

5x-2/x^2-2x-8 = 5x-2/(x-4)(x+2)

Then, 5x-2/(x-4)(x+2)= a/x-4 + b/x+2 ......eqn(i)

First put x= 4 in eqn(i) excluding the denominator where it is (x-4) ,

this implies 5(4)-2/4+2= a

so, a = 3

Similarly put x = -2 in eqn(i) ,

b= 5(-2)-2/-2-4

so, b= 2

Finally, 5x-2/(x-4)(x+2)= 3/x-4+ 2/x+2

Answer 2

Answer:

Required partial form is $\frac{120}{x + 2} + \frac{30}{x - 4}$

Step-by-step explanation:

Given,

$5x-\frac{ - 2}{ {x}^{2} - 2x - 8 } .......(1)$

Now,

${x}^{2} - 2x - 8$

We can solve it by middle term method,

${x}^{2} - (4 - 2)x - 8 \\ {x}^{2} - 4x + 2x - 8 \\ x(x - 4) + 2(x - 4) \\ ( x - 4)(x + 2)$

Now from the term (1),

$5x-\frac{ 2}{ (x+ 2)(x - 4)}$

Let,

$5x - \frac{2}{(x + 2)(x - 4)} = \frac{a}{x + 2} + \frac{b}{x - 4}$

Putting x=-2

Then,

$5 \times ( - 1) = \frac{b}{ - 2 - 4} \\ - 5 = - \frac{b}{6} \\ b = 30$

and putting x=4,

$5 \times 4 = \frac{a}{4 + 2} \\ 20 = \frac{a}{6} \\ a = 120$

So required partial fraction is $\frac{120}{x + 2} + \frac{30}{x - 4}$