Solve (2x^2-1)*(2x^2+1)-3x-(2x+1)=

$$( 2 x ^ { 2 } - 1 ) \cdot ( 2 x ^ { 2 } + 1 ) - 3 x - ( 2 x + 1 ) =$$

Evaluate

$$4x^{4}-5x-2$$

Solution Steps

Consider \left(2x^{2}-1\right)\left(2x^{2}+1\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 1.

$$\left(2x^{2}\right)^{2}-1-3x-\left(2x+1\right)$$

Expand \left(2x^{2}\right)^{2}.

$$2^{2}\left(x^{2}\right)^{2}-1-3x-\left(2x+1\right)$$

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

$$2^{2}x^{4}-1-3x-\left(2x+1\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{4}-1-3x-\left(2x+1\right)$$

To find the opposite of 2x+1, find the opposite of each term.

$$4x^{4}-1-3x-2x-1$$

Combine -3x and -2x to get -5x.

$$4x^{4}-1-5x-1$$

Subtract 1 from -1 to get -2.

$$4x^{4}-2-5x$$

Expand

$$4x^{4}-5x-2$$

Solution Steps

Consider \left(2x^{2}-1\right)\left(2x^{2}+1\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 1.

$$\left(2x^{2}\right)^{2}-1-3x-\left(2x+1\right)$$

Expand \left(2x^{2}\right)^{2}.

$$2^{2}\left(x^{2}\right)^{2}-1-3x-\left(2x+1\right)$$

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

$$2^{2}x^{4}-1-3x-\left(2x+1\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{4}-1-3x-\left(2x+1\right)$$

To find the opposite of 2x+1, find the opposite of each term.

$$4x^{4}-1-3x-2x-1$$

Combine -3x and -2x to get -5x.

$$4x^{4}-1-5x-1$$

Subtract 1 from -1 to get -2.

$$4x^{4}-2-5x$$