Question

Resolve into partial fractions

$\frac{7x − 1}{6 − 5x + x^2}$

Answer 1

(7x-1)/(x²-5x+6) = (7x-1)/{(x-2)(x-3)}
                       = A/(x-2) + B/(x-3)
                     = {(A+B)x - (3A+2B)}/{(x-2)(x-3)}
compare the coefficient of power of x 
A+B = 7 ----------(1)
3A+2B = 1 --------------(2)
on solving the both equation we get ,
A=-13   B=20
hence partial fraction is
 = -13/(x-2) + 20/(x-3)

Answer 2

The partial fractions are, $\frac{-15}{2-x} +\frac{22}{3-x} .$

• A complex rational statement can be broken down into two or more simpler fractions to generate partial fractions. Since fractions with algebraic expressions are typically challenging to solve, we divide them up into a large number of smaller fractions using the principles of partial fractions.
• During decomposition, the denominator is often an algebraic expression that is factorised to make it easier to create partial fractions. A partial fraction is the opposite of how rational expressions are added.

Here, according to the given information, we are given that,

$\frac{7x-1}{6-5x+x^{2} }$

Now, resolving the denominator into factors, we get,

$\frac{7x-1}{6-5x+x^{2} }\\=\frac{7x-1}{6-2x-3x+x^{2} } \\=\frac{7x-1}{2(3-x)-x(3-x)} \\=\frac{7x-1}{(2-x)(3-x)}$

Now, Let this be re-written as,

$\frac{A}{2-x}+\frac{B}{3-x} \\=\frac{A(3-x)+B(2-x)}{(2-x)(3-x)} \\=\frac{(-A-B)x+(3A+2B)}{(2-x)(3-x)}$

Comparing the coefficients, we get,

$-A-B = 7.$

Also, $3A+2B = -1.$

Then, we have, $B = 7-A.$

$3A+2(7-A)=-1.$

Or, A = -15.

B = 22.

Then, the partial fractions are,

$\frac{-15}{2-x} +\frac{22}{3-x} .$

Hence, the partial fractions are, $\frac{-15}{2-x} +\frac{22}{3-x} .$

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