Solve (4x^2-sqrt{6})^2

$$( 4 x ^ { 2 } - \sqrt { 6 } ) ^ { 2 }$$

Expand

$$16x^{4}-8\sqrt{6}x^{2}+6$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x^{2}-\sqrt{6}\right)^{2}.

$$16\left(x^{2}\right)^{2}-8x^{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$16x^{4}-8x^{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$16x^{4}-8x^{2}\sqrt{6}+6$$

Evaluate

$$\left(4x^{2}-\sqrt{6}\right)^{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4x^{2}-\sqrt{6}\right)^{2}.

$$16\left(x^{2}\right)^{2}-8x^{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$16x^{4}-8x^{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$16x^{4}-8x^{2}\sqrt{6}+6$$