Express \[\frac{9x^{2} + 20x - 10}{(x + 2)(3x - 1)}}\] in partial fractions. - Edexcel - A-Level Maths Pure - Question 15 - 2013 - Paper 1

To express the given rational function in partial fractions, we can set it up as follows:

[\frac{9x^{2} + 20x - 10}{(x + 2)(3x - 1)} = \frac{A}{x + 2} + \frac{B}{3x - 1}]

Next, we multiply both sides by the denominator ((x + 2)(3x - 1)) to eliminate the fraction:

[9x^{2} + 20x - 10 = A(3x - 1) + B(x + 2)]

Now, let's expand the right side:

[A(3x - 1) = 3Ax - A] [B(x + 2) = Bx + 2B]

Combining these gives us:

[9x^{2} + 20x - 10 = (3A + B)x + (-A + 2B)]

To find coefficients, we can equate the coefficients of corresponding powers of x:

1. For the coefficient of (x): [3A + B = 20 ]
2. For the constant term: [-A + 2B = -10]

Now we have a system of equations:

1. ( 3A + B = 20 ) (Equation 1)
2. ( -A + 2B = -10) (Equation 2)

To solve this system, we can express (B) from Equation 1:

[B = 20 - 3A]

Substituting this into Equation 2 gives us:

[-A + 2(20 - 3A) = -10 ] [-A + 40 - 6A = -10] [ -7A + 40 = -10] [-7A = -50] [A = \frac{50}{7}]

Substituting back to find (B):

[B = 20 - 3 \times \frac{50}{7} = 20 - \frac{150}{7} = \frac{140}{7} - \frac{150}{7} = -\frac{10}{7}]

Now we substitute the values of (A) and (B) back into our partial fractions:

[\frac{9x^{2} + 20x - 10}{(x + 2)(3x - 1)} = \frac{\frac{50}{7}}{x + 2} + \frac{-\frac{10}{7}}{3x - 1}]

Thus, the final answer is:

[\frac{9x^{2} + 20x - 10}{(x + 2)(3x - 1)} = \frac{50}{7(x + 2)} - \frac{10}{7(3x - 1)}]