How do you simplify $\frac{x}{x^{2} - 16} + \frac{x - 2}{x^{2} - 5 x + 4}$ $x/(x^2-16) + (x-2)/(x^2-5x+4)$ ?

Question

1782 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-15T08:02:07

$\frac{2 x^{2} + x - 8}{(x - 4) (x + 4) (x - 1)}$ $(2x^2+x-8)/((x-4)(x+4)(x-1))$

Note that

$x^{2} - 16 = (x - 4) (x + 4)$ $x^2-16=(x-4)(x+4)$

$x^{2} - 5 x + 4 = (x - 4) (x - 1)$ $x^2-5x+4=(x-4)(x-1)$
so we have to add

$\frac{x (x - 1)}{(x - 4) (x + 4) (x - 1)}$ $(x(x-1))/((x-4)(x+4)(x-1))$

$\frac{(x - 2) (x + 4)}{(x - 4) (x - 1) (x + 4)}$ $((x-2)(x+4))/((x-4)(x-1)(x+4))$
and we have

$x (x - 1) = x^{2} - x$ $x(x-1)=x^2-x$

$(x - 2) (x + 4) = x^{2} + 2 x - 8$ $(x-2)(x+4)=x^2+2x-8$
so we get the result above.

How do you simplify x/(x^2-16) + (x-2)/(x^2-5x+4)xx2−16+x−2x2−5x+4x/(x^2-16) + (x-2)/(x^2-5x+4)?