Solve [(2x^2-x)+3][(2x^2-x)-3]

$$[ ( 2 x ^ { 2 } - x ) + 3 ] [ ( 2 x ^ { 2 } - x ) - 3 ]$$

Evaluate

$$4x^{4}-4x^{3}+x^{2}-9$$

Solution Steps

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

$$\left(2x^{2}-x\right)^{2}-9$$

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

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

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

$$4x^{4}-4x^{2}x+x^{2}-9$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$4x^{4}-4x^{3}+x^{2}-9$$

Expand

$$4x^{4}-4x^{3}+x^{2}-9$$

Solution Steps

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

$$\left(2x^{2}-x\right)^{2}-9$$

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

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

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

$$4x^{4}-4x^{2}x+x^{2}-9$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$4x^{4}-4x^{3}+x^{2}-9$$