Solve (3sqrt{2}+2)(3sqrt{2}-2)+(2-275)^2=

$$( 3 \sqrt { 2 } + 2 ) ( 3 \sqrt { 2 } - 2 ) + ( 2 - 275 ) ^ { 2 } =$$

Evaluate

$$74543$$

Solution Steps

Consider \left(3\sqrt{2}+2\right)\left(3\sqrt{2}-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 2.

$$\left(3\sqrt{2}\right)^{2}-4+\left(2-275\right)^{2}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$3^{2}\left(\sqrt{2}\right)^{2}-4+\left(2-275\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{2}\right)^{2}-4+\left(2-275\right)^{2}$$

The square of \sqrt{2} is 2.

$$9\times 2-4+\left(2-275\right)^{2}$$

Multiply 9 and 2 to get 18.

$$18-4+\left(2-275\right)^{2}$$

Subtract 4 from 18 to get 14.

$$14+\left(2-275\right)^{2}$$

Subtract 275 from 2 to get -273.

$$14+\left(-273\right)^{2}$$

Calculate -273 to the power of 2 and get 74529.

$$14+74529$$

Add 14 and 74529 to get 74543.

$$74543$$

Factor

$$7\times 23\times 463$$