The evaluation of the algebraic fraction x+2x+1/x^2-25 is[tex]\mathbf{\implies -\dfrac{3}{5(x+5)} +\dfrac{18}{5(x-5)}}[/tex]
The division of an algebraic expression can be carried out by using the long division method or expansion of the.
Given that:
[tex]\mathbf{\dfrac{x+2x+21}{x^2-25}}[/tex]
where:
By expanding the denominator, we have:
[tex]\mathbf{\dfrac{x+2x+21}{(x+5)(x-5)}}[/tex]
[tex]\mathbf{\dfrac{3x+21}{(x+5)(x-5)}}[/tex]
Creating a partial fraction from the denominator, we have:
[tex]\mathbf{\dfrac{3x+21}{(x+5)(x-5)} = \dfrac{a_o}{x+5} +\dfrac{a_1}{x-5}}[/tex]
Multiply the equation by the denominator, and we have:
[tex]\mathbf{\dfrac{(3x+21)(x+5)+(x-5)}{(x+5)(x-5)} = \dfrac{a_o(x+5)(x-5)}{x+5} +\dfrac{a_1(x+5)(x-5)}{x-5}}[/tex]
Simplifying, we have:
3x + 21 = a₀(x-5)+a₁(x+5)
Solve for the unknown parameter by plugging the real roots of the denominator: -5, 5
[tex]\mathbf{a_o =\dfrac{-3}{5}}[/tex]
[tex]\mathbf{a_1 =\dfrac{18}{5}}[/tex]
Replacing the solutions into the partial fractions parameters to obtain the final result, we have:
[tex]\mathbf{\dfrac{-(\dfrac{3}{5})}{x+5} +\dfrac{\dfrac{18}{5}}{x-5}}[/tex]
[tex]\mathbf{\implies -\dfrac{3}{5(x+5)} +\dfrac{18}{5(x-5)}}[/tex]
Learn more about the division of algebraic fractions here:
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