Solve (x-2)(x+2)(x^2+4)=x^2+b

$$( x - 2 ) ( x + 2 ) ( x ^ { 2 } + 4 ) = x ^ { 2 } + b$$

Solve for b

$$b=x^{4}-x^{2}-16$$

Solution Steps

Use the distributive property to multiply x-2 by x+2 and combine like terms.

$$\left(x^{2}-4\right)\left(x^{2}+4\right)=x^{2}+b$$

Consider \left(x^{2}-4\right)\left(x^{2}+4\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 4.

$$\left(x^{2}\right)^{2}-16=x^{2}+b$$

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

$$x^{4}-16=x^{2}+b$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}+b=x^{4}-16$$

Subtract x^{2} from both sides.

$$b=x^{4}-16-x^{2}$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{-2\sqrt{4b+65}+2}}{2}$$

$$x=\frac{\sqrt{-2\sqrt{4b+65}+2}}{2}$$

$$x=\frac{\sqrt{2\sqrt{4b+65}+2}}{2}$$

$$x=-\frac{\sqrt{2\sqrt{4b+65}+2}}{2}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{-2\sqrt{4b+65}+2}}{2}\text{; }x=-\frac{\sqrt{-2\sqrt{4b+65}+2}}{2}\text{, }&b\geq -\frac{65}{4}\text{ and }b\leq -16\\x=\frac{\sqrt{2\sqrt{4b+65}+2}}{2}\text{; }x=-\frac{\sqrt{2\sqrt{4b+65}+2}}{2}\text{, }&b\geq -\frac{65}{4}\end{matrix}\right.$$