# `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)` .**