Resolve into partial fractions 7-5x/2x^2+x-1

1 answer:

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The decomposition of partial fractions is to start with the simplified reply and then take it apart, to "decompose" the final expression into its initial polynomial fractions.

Given:

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

To find:

partial fractions=?

Solution:

$\to \bold{\frac{7-5x}{2x^2+x-1}}\\\\\to \bold{\frac{7-5x}{2x^2+x(2-1)-1}}\\\\\to \bold{\frac{7-5x}{2x^2+2x-x-1}}\\\\\to \bold{\frac{7-5x}{2x(x+1)-1(x+1)}}\\\\\to \bold{\frac{7-5x}{(2x-1)(x+1)} = \frac{A}{(2x-1)} - \frac{B}{(x+1)} }\\\\\to \bold{7-5x = \frac{A ((2x-1)(x+1))}{(2x-1)} - \frac{B((2x-1)(x+1))}{(x+1)} }\\\\\to \bold{7-5x = A(x+1) -B(2x-1) }\\\\$

putting $x=-1$

$\to \bold{7-5(-1) = A(-1+1) -B(2(-1)-1) }\\\\\to \bold{7+5 = A(0) -B(-2-1) }\\\\\to \bold{12 = +3B }\\\\\to \bold{B = \frac{12}{3} }\\\\\to \bold{B = 4 }\\\\$

putting $x= \frac{1}{2}$

$\to \bold{7-5(\frac{1}{2}) = A(\frac{1}{2}+1) -B(2(\frac{1}{2})-1) }\\\\\to \bold{7-\frac{5}{2} = A(\frac{3}{2}) -B((\frac{2}{2})-1) }\\\\\to \bold{7-\frac{5}{2} = A(\frac{3}{2}) -B(1-1) }\\\\\to \bold{\frac{14-5}{2} = A(\frac{3}{2}) -B(0) }\\\\\to \bold{\frac{9}{2} = A(\frac{3}{2})}\\\\\to \bold{\frac{9}{2} \times \frac{2}{3} = A}\\\\\to \bold{\frac{9}{3} = A}\\\\\to \bold{A=3}\\\\$