Solve 3(sqrt{2}+2)^2-12(sqrt{2}+2)+6

$$3 ( \sqrt { 2 } + 2 ) ^ { 2 } - 12 ( \sqrt { 2 } + 2 ) + 6$$

Evaluate

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Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+2\right)^{2}.

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

The square of \sqrt{2} is 2.

$$3\left(2+4\sqrt{2}+4\right)-12\left(\sqrt{2}+2\right)+6$$

Add 2 and 4 to get 6.

$$3\left(6+4\sqrt{2}\right)-12\left(\sqrt{2}+2\right)+6$$

Use the distributive property to multiply 3 by 6+4\sqrt{2}.

$$18+12\sqrt{2}-12\left(\sqrt{2}+2\right)+6$$

Use the distributive property to multiply -12 by \sqrt{2}+2.

$$18+12\sqrt{2}-12\sqrt{2}-24+6$$

Combine 12\sqrt{2} and -12\sqrt{2} to get 0.

$$18-24+6$$

Subtract 24 from 18 to get -6.

$$-6+6$$

Add -6 and 6 to get 0.

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Factor

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