Algebraic Simplification

In the following examples and problems, the term "simplify" indicates to eliminate compound fractions, factor as much as possible, put terms over a common denominator when feasible, and avoid negative exponents.

Sol A

$$\frac{x^3-8}{x^2-4}$$

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

$$=\frac{x^2+2x+4}{x+2}$$

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Sol B

$$\frac{1/x^2-1/9}{x-3}$$

$$=\frac{9-x^2}{(x-3)(9x^2)}$$

(multiplying the top and bottom by $9x^2$)

$$=\frac{-(x^2-9)}{(x-3)(9x^2)}$$

$$=\frac{-(x-3)(x+3)}{(x-3)(9x^2)}$$

$$=\frac{-(x+3)}{9x^2}$$

Sol C

$$\frac{\sqrt{x+h}-\sqrt{x}}{h}$$

(multiplying the top and bottom by $\sqrt{x+h}+\sqrt{x}$)

$$=\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}$$

$$=\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$$

$$=\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

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Sol 1

$$5/3x^{2/3}-10/3x^{-1/3}=(5/3)x^{-1/3}(x-2)$$

$$\frac{(5/3)(x-2)}{x^{1/3}}=\frac{5(x-2)}{3x^{1/3}}$$

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Sol 2

$$x(1/2)(2x+3)^{-1/2}(2)+\sqrt{2x+3}$$

$$=(2x+3)^{-1/2}[x+(2x+3)]=\frac{3x+3}{(2x+3)^{1/2}}$$

$$=\frac{3(x+1)}{\sqrt{2x+3}}$$

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Sol 3

$$t^2(1/2)(t-2)^{-1/2}+2t\sqrt{t-2}$$

$$=(1/2)t(t-2)^{-1/2}[t+4(t-2)]$$

$$=\frac{(1/2)t[5t-8]}{(t-2)^{1/2}}$$

$$=\frac{t(5t-8)}{2\sqrt{t-2}}$$

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Sol 4

$$\sqrt{x}(2)(x-2)+(1/2)x^{-1/2}(x-2)^2$$

$$=(1/2)x^{-1/2}(x-2)[4x+(x-2)]=\frac{(x-2)(5x-2)}{2x^{1/2}}$$

$$=\frac{(x-2)(5x-2)}{2\sqrt{x}}$$

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Sol 5

$$x(1/3)(x^2+5)^{-2/3}(2x)+(x^2+5)^{1/3}$$

$$=(1/3)(x^2+5)^{-2/3}[2x^2+3(x^2+5)]$$

$$=\frac{(1/3)[5x^2+15]}{(x^2+5)^{2/3}}$$

$$=\frac{5(x^2+3)}{3(x^2+5)^{2/3}}$$

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Sol 6

$$\frac{\sqrt{25+x^2}-x(1/2)(25+x^2)^{-1/2}(2x)}{25+x^2}$$

$$=\frac{\sqrt{25+x^2}-\frac{x^2}{(25+x^2)^{1/2}}}{25+x^2}$$

(multiplying the top and bottom by $(25+x^2)^{1/2}$)

$$=\frac{(25+x^2)-x^2}{(25+x^2)(25+x^2)^{1/2}}$$

$$=\frac{25}{(25+x^2)^{3/2}}$$

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Sol 7

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

$$=\frac{x^2(x^2+1)^{-1/2}-\sqrt{x^2+1}}{x^2}$$

$$=\frac{\frac{x^2}{\sqrt{x^2+1}}-\sqrt{x^2+1}}{x^2}$$

$$=\frac{x^2-(x^2+1)}{x^2\sqrt{x^2+1}}$$

$$=\frac{-1}{x^2\sqrt{x^2+1}}=-\frac{1}{x^2\sqrt{x^2+1}}$$

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Sol 8

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

$$=\frac{(2x)(x^2+1)[-(x^2+1)-2(1-x^2)]}{(x^2+1)^4}$$

$$=\frac{(2x)[-x^2-1-2+2x^2]}{(x^2+1)^3}$$

$$=\frac{(2x)[x^2-3]}{(x^2+1)^3}$$

$$=\frac{(2x)(x-\sqrt{3})(x+\sqrt{3})}{(x^2+1)^3}$$

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