Solve (x+1+frac{sqrt{7}}{2sqrt{2}})(x+1-frac{sqrt{7}}{2sqrt{2}})-(x^2+2x+1/8)

$$(x+1+ \frac{ \sqrt{ 7 } }{ 2 \sqrt{ 2 } } )(x+1- \frac{ \sqrt{ 7 } }{ 2 \sqrt{ 2 } } )-( { x }^{ 2 } +2x+ \frac{ 1 }{ 8 } )$$

Evaluate

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Short Solution Steps

Rationalize the denominator of \frac{\sqrt{7}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\left(x+1+\frac{\sqrt{7}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

The square of \sqrt{2} is 2.

$$\left(x+1+\frac{\sqrt{7}\sqrt{2}}{2\times 2}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\left(x+1+\frac{\sqrt{14}}{2\times 2}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

Multiply 2 and 2 to get 4.

$$\left(x+1+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x+1 times \frac{4}{4}.

$$\left(\frac{4\left(x+1\right)}{4}+\frac{\sqrt{14}}{4}\right)\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

Since \frac{4\left(x+1\right)}{4} and \frac{\sqrt{14}}{4} have the same denominator, add them by adding their numerators.

$$\frac{4\left(x+1\right)+\sqrt{14}}{4}\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

Do the multiplications in 4\left(x+1\right)+\sqrt{14}.

$$\frac{4x+4+\sqrt{14}}{4}\left(x+1-\frac{\sqrt{7}}{2\sqrt{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

Rationalize the denominator of \frac{\sqrt{7}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{4x+4+\sqrt{14}}{4}\left(x+1-\frac{\sqrt{7}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

The square of \sqrt{2} is 2.

$$\frac{4x+4+\sqrt{14}}{4}\left(x+1-\frac{\sqrt{7}\sqrt{2}}{2\times 2}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{4x+4+\sqrt{14}}{4}\left(x+1-\frac{\sqrt{14}}{2\times 2}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

Multiply 2 and 2 to get 4.

$$\frac{4x+4+\sqrt{14}}{4}\left(x+1-\frac{\sqrt{14}}{4}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x+1 times \frac{4}{4}.

$$\frac{4x+4+\sqrt{14}}{4}\left(\frac{4\left(x+1\right)}{4}-\frac{\sqrt{14}}{4}\right)-\left(x^{2}+2x+\frac{1}{8}\right)$$

Since \frac{4\left(x+1\right)}{4} and \frac{\sqrt{14}}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4x+4+\sqrt{14}}{4}\times \frac{4\left(x+1\right)-\sqrt{14}}{4}-\left(x^{2}+2x+\frac{1}{8}\right)$$

Do the multiplications in 4\left(x+1\right)-\sqrt{14}.

$$\frac{4x+4+\sqrt{14}}{4}\times \frac{4x+4-\sqrt{14}}{4}-\left(x^{2}+2x+\frac{1}{8}\right)$$

Multiply \frac{4x+4+\sqrt{14}}{4} times \frac{4x+4-\sqrt{14}}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(4x+4+\sqrt{14}\right)\left(4x+4-\sqrt{14}\right)}{4\times 4}-\left(x^{2}+2x+\frac{1}{8}\right)$$

Multiply 4 and 4 to get 16.

$$\frac{\left(4x+4+\sqrt{14}\right)\left(4x+4-\sqrt{14}\right)}{16}-\left(x^{2}+2x+\frac{1}{8}\right)$$

Use the distributive property to multiply 4x+4+\sqrt{14} by 4x+4-\sqrt{14} and combine like terms.

$$\frac{16x^{2}+32x+16-\left(\sqrt{14}\right)^{2}}{16}-\left(x^{2}+2x+\frac{1}{8}\right)$$

The square of \sqrt{14} is 14.

$$\frac{16x^{2}+32x+16-14}{16}-\left(x^{2}+2x+\frac{1}{8}\right)$$

Subtract 14 from 16 to get 2.

$$\frac{16x^{2}+32x+2}{16}-\left(x^{2}+2x+\frac{1}{8}\right)$$

To find the opposite of x^{2}+2x+\frac{1}{8}, find the opposite of each term.

$$\frac{16x^{2}+32x+2}{16}-x^{2}-2x-\frac{1}{8}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 16 and 8 is 16. Multiply \frac{1}{8} times \frac{2}{2}.

$$\frac{16x^{2}+32x+2}{16}-x^{2}-2x-\frac{2}{16}$$

Since \frac{16x^{2}+32x+2}{16} and \frac{2}{16} have the same denominator, subtract them by subtracting their numerators.

$$\frac{16x^{2}+32x+2-2}{16}-x^{2}-2x$$

Combine like terms in 16x^{2}+32x+2-2.

$$\frac{16x^{2}+32x}{16}-x^{2}-2x$$

Divide each term of 16x^{2}+32x by 16 to get 2x+x^{2}.

$$2x+x^{2}-x^{2}-2x$$

Combine x^{2} and -x^{2} to get 0.

$$2x-2x$$

Combine 2x and -2x to get 0.

$$0$$

Factor

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