Solve (1)/((x)^(2)+2x-24)+(2)/((x)^(2)-2x-8)-(3)/((x)^(2)+8x+12) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{ 1 }{ { x }^{ 2 } +2x-24 } + \frac{ 2 }{ { x }^{ 2 } -2x-8 } - \frac{ 3 }{ { x }^{ 2 } +8x+12 }$$

Evaluate

$\frac{26}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$

Short Solution Steps

Factor $x^{2}+2x-24$. Factor $x^{2}-2x-8$.

$$\frac{1}{\left(x-4\right)\left(x+6\right)}+\frac{2}{\left(x-4\right)\left(x+2\right)}-\frac{3}{x^{2}+8x+12}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $\left(x-4\right)\left(x+6\right)$ and $\left(x-4\right)\left(x+2\right)$ is $\left(x-4\right)\left(x+2\right)\left(x+6\right)$. Multiply $\frac{1}{\left(x-4\right)\left(x+6\right)}$ times $\frac{x+2}{x+2}$. Multiply $\frac{2}{\left(x-4\right)\left(x+2\right)}$ times $\frac{x+6}{x+6}$.

$$\frac{x+2}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}+\frac{2\left(x+6\right)}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}-\frac{3}{x^{2}+8x+12}$$

Since $\frac{x+2}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$ and $\frac{2\left(x+6\right)}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$ have the same denominator, add them by adding their numerators.

$$\frac{x+2+2\left(x+6\right)}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}-\frac{3}{x^{2}+8x+12}$$

Do the multiplications in $x+2+2\left(x+6\right)$.

$$\frac{x+2+2x+12}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}-\frac{3}{x^{2}+8x+12}$$

Combine like terms in $x+2+2x+12$.

$$\frac{3x+14}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}-\frac{3}{x^{2}+8x+12}$$

Factor $x^{2}+8x+12$.

$$\frac{3x+14}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}-\frac{3}{\left(x+2\right)\left(x+6\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $\left(x-4\right)\left(x+2\right)\left(x+6\right)$ and $\left(x+2\right)\left(x+6\right)$ is $\left(x-4\right)\left(x+2\right)\left(x+6\right)$. Multiply $\frac{3}{\left(x+2\right)\left(x+6\right)}$ times $\frac{x-4}{x-4}$.

$$\frac{3x+14}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}-\frac{3\left(x-4\right)}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$$

Since $\frac{3x+14}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$ and $\frac{3\left(x-4\right)}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{3x+14-3\left(x-4\right)}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$$

Do the multiplications in $3x+14-3\left(x-4\right)$.

$$\frac{3x+14-3x+12}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$$

Combine like terms in $3x+14-3x+12$.

$$\frac{26}{\left(x-4\right)\left(x+2\right)\left(x+6\right)}$$

Expand $\left(x-4\right)\left(x+2\right)\left(x+6\right)$.

$$\frac{26}{x^{3}+4x^{2}-20x-48}$$

Differentiate w.r.t. x

$-\frac{26\left(\left(3x+4\right)^{2}-76\right)}{3\left(\left(x-4\right)\left(x+2\right)\left(x+6\right)\right)^{2}}$