How do you simplify $\frac{a + 2 b}{x + a} - \frac{a - 2 b}{x - a} - \frac{4 b x - 2 a^{2}}{x^{2} - a^{2}}$ $(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)$ ?

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How do you simplify $\frac{a + 2 b}{x + a} - \frac{a - 2 b}{x - a} - \frac{4 b x - 2 a^{2}}{x^{2} - a^{2}}$ $(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)$ ?

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Answer 1 · 2017-08-15T14:38:14

$= 0$ $=0$

Explanation:

$\frac{a + 2 b}{x + a} - \frac{a - 2 b}{x - a} - \frac{4 b x - 2 a^{2}}{x^{2} - a^{2}} \leftarrow$ $(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)" "larr$ factorise

$\frac{a + 2 b}{x + a} - \frac{a - 2 b}{x - a} - \frac{(4 b x - 2 a^{2})}{(x + a) (x - a)}$ $(a+2b)/(x+a)-(a-2b)/(x-a)-((4bx-2a^2))/((x+a)(x-a))$

Find a common denominator and equivalent fractions:

$\frac{a + 2 b}{x + a} \times \frac{x - a}{x - a} - \frac{a - 2 b}{x - a} \times \frac{x + a}{x + a} - \frac{(4 b x - 2 a^{2})}{(x + a) (x - a)}$ $(a+2b)/(x+a)xxcolor(blue)((x-a)/(x-a))-(a-2b)/(x-a)xxcolor(blue)( (x+a)/(x+a))-((4bx-2a^2))/((x+a)(x-a))$

$\frac{(a + 2 b) (x - a) - (a - 2 b) (x + a) - (4 b x - 2 a^{2})}{(x + a) (x - a)}$ $((a+2b)(x-a) -(a-2b)(x+a) - (4bx-2a^2))/((x+a)(x-a))$

$= \frac{(a x - a^{2} + 2 b x - 2 a b) - (a x + a^{2} - 2 b x - 2 a b) - (4 b x - 2 a^{2})}{(x + a) (x - a)}$ $=((ax-a^2+2bx-2ab)-(ax+a^2-2bx-2ab)-(4bx-2a^2))/((x+a)(x-a))$

$= \frac{a x - a^{2} + 2 b x - 2 a b - a x - a^{2} + 2 b x + 2 a b - 4 b x + 2 a^{2}}{(x + a) (x - a)}$ $=(color(blue)(ax)color(red)(-a^2)color(forestgreen)(+2bx)color(magenta)(-2ab)color(blue)(-ax)color(red)(-a^2)color(forestgreen)(+2bx)color(magenta)(+2ab)color(forestgreen)(-4bx)color(red)(+2a^2))/((x+a)(x-a))$

$= \frac{0 a x + 0 a^{2} + 0 b x + 0 a b}{(x + a) (x - a)}$ $=(color(blue)(0ax)color(red)(+0a^2)color(forestgreen)(+0bx)color(magenta)(+0ab))/((x+a)(x-a))$

$= 0$ $=0$

Answer 2 · 2017-08-15T15:11:01

$0$ $0$

Explanation:

$before we can add/subtract fractions we require them to$ $"before we can add/subtract fractions we require them to"$
$have a common denominator$ $"have a "color(blue)"common denominator"$

$x^{2} - a^{2} is a difference of squares$ $x^2-a^2" is a "color(blue)"difference of squares"$

$\Rightarrow a^{2} - x^{2} = (x - a) (x + a)$ $rArra^2-x^2=(x-a)(x+a)$

$multiply numerator/denominator of \frac{a + 2 b}{x + a} by (x - a)$ $"multiply numerator/denominator of "(a+2b)/(x+a)" by "(x-a)$

$multiply numerator/denominator of \frac{a - 2 b}{x - a} by (x + a)$ $"multiply numerator/denominator of "(a-2b)/(x-a)" by "(x+a)$

$\frac{(a + 2 b) (x - a)}{(x - a) (x + a)} - \frac{(a - 2 b) (x + a)}{(x - a) (x + a)} - \frac{4 b x - 2 a^{2}}{(x - a) (x + a)}$ $((a+2b)(x-a))/((x-a)(x+a))-((a-2b)(x+a))/((x-a)(x+a))-(4bx-2a^2)/((x-a)(x+a))$

$now add/subtract the numerators leaving the denominator$ $"now add/subtract the numerators leaving the denominator"$
$first multiplying out the brackets$ $"first multiplying out the brackets"$

$\frac{a x + 2 b x - a^{2} - 2 a b - (a x - 2 b x + a^{2} - 2 a b) - (4 b x - 2 a^{2})}{(x - a) (x + a)}$ $(ax+2bx-a^2-2ab-(ax-2bx+a^2-2ab)-(4bx-2a^2))/((x-a)(x+a))$

$= \frac{a x + 2 b x - a^{2} - 2 a b - a x + 2 b x - a^{2} + 2 a b - 4 b x + 2 a^{2}}{(x - a) (x + a)}$ $=(ax+2bx-a^2-2ab-ax+2bx-a^2+2ab-4bx+2a^2)/((x-a)(x+a))$

$= 0$ $=0$

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How do you simplify $\frac{a + 2 b}{x + a} - \frac{a - 2 b}{x - a} - \frac{4 b x - 2 a^{2}}{x^{2} - a^{2}}$ $(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)$ ?

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Explanation:

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How do you simplify a+2bx+a−a−2bx−a−4bx−2a2x2−a2(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)?

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How do you simplify $\frac{a + 2 b}{x + a} - \frac{a - 2 b}{x - a} - \frac{4 b x - 2 a^{2}}{x^{2} - a^{2}}$ $(a+2b)/(x+a)-(a-2b)/(x-a)-(4bx-2a^2)/(x^2-a^2)$ ?