Solve 142x+1/(x-1)^2(x^2+1)+7

$$14 \frac { 2 x + 1 } { ( x - 1 ) ^ { 2 } ( x ^ { 2 } + 1 ) } + 7$$

Evaluate

$$\frac{7\left(x^{4}-2x^{3}+2x^{2}+2x+3\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Short Solution Steps

Use the distributive property to multiply \left(x-1\right)^{2} by x^{2}+1.

$$14\times \frac{2x+1}{\left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}}+7$$

Express 14\times \frac{2x+1}{\left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}} as a single fraction.

$$\frac{14\left(2x+1\right)}{\left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}}+7$$

Factor \left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}.

$$\frac{14\left(2x+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}+7$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7 times \frac{\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}.

$$\frac{14\left(2x+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}+\frac{7\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Since \frac{14\left(2x+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)} and \frac{7\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{14\left(2x+1\right)+7\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Do the multiplications in 14\left(2x+1\right)+7\left(x-1\right)^{2}\left(x^{2}+1\right).

$$\frac{28x+14+7x^{4}+7x^{2}-14x^{3}-14x+7x^{2}+7}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Combine like terms in 28x+14+7x^{4}+7x^{2}-14x^{3}-14x+7x^{2}+7.

$$\frac{14x+21+7x^{4}+14x^{2}-14x^{3}}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Expand \left(x-1\right)^{2}\left(x^{2}+1\right).

$$\frac{14x+21+7x^{4}+14x^{2}-14x^{3}}{x^{4}-2x^{3}+2x^{2}-2x+1}$$

Expand

$$\frac{7\left(x^{4}-2x^{3}+2x^{2}+2x+3\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Short Solution Steps

Use the distributive property to multiply \left(x-1\right)^{2} by x^{2}+1.

$$14\times \frac{2x+1}{\left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}}+7$$

Express 14\times \frac{2x+1}{\left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}} as a single fraction.

$$\frac{14\left(2x+1\right)}{\left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}}+7$$

Factor \left(x-1\right)^{2}x^{2}+\left(x-1\right)^{2}.

$$\frac{14\left(2x+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}+7$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7 times \frac{\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}.

$$\frac{14\left(2x+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}+\frac{7\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Since \frac{14\left(2x+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)} and \frac{7\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{14\left(2x+1\right)+7\left(x-1\right)^{2}\left(x^{2}+1\right)}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Do the multiplications in 14\left(2x+1\right)+7\left(x-1\right)^{2}\left(x^{2}+1\right).

$$\frac{28x+14+7x^{4}+7x^{2}-14x^{3}-14x+7x^{2}+7}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Combine like terms in 28x+14+7x^{4}+7x^{2}-14x^{3}-14x+7x^{2}+7.

$$\frac{14x+21+7x^{4}+14x^{2}-14x^{3}}{\left(x-1\right)^{2}\left(x^{2}+1\right)}$$

Expand \left(x-1\right)^{2}\left(x^{2}+1\right).

$$\frac{14x+21+7x^{4}+14x^{2}-14x^{3}}{x^{4}-2x^{3}+2x^{2}-2x+1}$$