`1/(2x^2 + x)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.
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`1/(2x^2+x)`
To decompose this into partial fractions, factor the denominator.
`1/(x(2x+1))`
Write a fraction for each factor. Since the numerators are still unknown, assign a variable to each numerator.
`A/x and B/(2x+1)`
Add these two fractions and set it equal to the given fraction.
`1/(x(2x+1)) = A/x + B/(2x+1)`
To solve for the values of A and B, eliminate the fractions in the equation. So, multiply both sides by the LCD.
`x(2x+1)*1/(x(2x+1)) = (A/x + B/(2x+1))*x(2x+1)`
`1=A(2x+1) + Bx`
Then, plug-in the roots of the factors.
For the factor 2x + 1, its root is x=-1/2.
`1=A(2*(-1/2)+1) + B(-1/2)`
`1=A(-1+1)+B(-1/2)`
`1=-1/2B`
`-2=B`
For the factor x, its root is x=0.
`1=A(2x+1)+Bx`
`1=A(2*0+1)+B*0`
`1=A`
So the partial fraction decomposition of the rational expression is:
`1/x + (-2)/(2x+1)`
And the sign before the second fraction simplifies to:
`1/x - 2/(2x+1)`
To check, express these two fractions with same denominators.
`1/x-2/(2x+1) = 1/x*(2x+1)/(2x+1) - 2/(2x+1)*x/x = (2x+1)/(x(2x+1)) - (2x)/(x(2x+1))`
Now that they have same denominators, proceed to subtract them.
`=(2x+1-2x)/(x(2x+1)) = 1/(x(2x+1)) = 1/(2x^2+x)`
Therefore, `1/(2x^2+x) = 1/x-2/(2x+1)` .
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