Solve left(x^2+3xright)^2-left(2x+1right)^2

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

Evaluate

$$x^{4}+6x^{3}+5x^{2}-4x-1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+3x\right)^{2}.

$$\left(x^{2}\right)^{2}+6x^{2}x+9x^{2}-\left(2x+1\right)^{2}$$

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

$$x^{4}+6x^{2}x+9x^{2}-\left(2x+1\right)^{2}$$

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

$$x^{4}+6x^{3}+9x^{2}-\left(2x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$x^{4}+6x^{3}+9x^{2}-\left(4x^{2}+4x+1\right)$$

To find the opposite of 4x^{2}+4x+1, find the opposite of each term.

$$x^{4}+6x^{3}+9x^{2}-4x^{2}-4x-1$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$x^{4}+6x^{3}+5x^{2}-4x-1$$

Expand

$$x^{4}+6x^{3}+5x^{2}-4x-1$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x^{2}+3x\right)^{2}.

$$\left(x^{2}\right)^{2}+6x^{2}x+9x^{2}-\left(2x+1\right)^{2}$$

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

$$x^{4}+6x^{2}x+9x^{2}-\left(2x+1\right)^{2}$$

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

$$x^{4}+6x^{3}+9x^{2}-\left(2x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$x^{4}+6x^{3}+9x^{2}-\left(4x^{2}+4x+1\right)$$

To find the opposite of 4x^{2}+4x+1, find the opposite of each term.

$$x^{4}+6x^{3}+9x^{2}-4x^{2}-4x-1$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$x^{4}+6x^{3}+5x^{2}-4x-1$$