Solve (x-2q)^2-4*(x+q)^2=

$$( x - 2 q ) ^ { 2 } - 4 \cdot ( x + q ) ^ { 2 } =$$

Evaluate

$$-3x\left(x+4q\right)$$

Solution Steps

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

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

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

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

Use the distributive property to multiply -4 by x^{2}+2xq+q^{2}.

$$x^{2}-4xq+4q^{2}-4x^{2}-8xq-4q^{2}$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-4xq+4q^{2}-8xq-4q^{2}$$

Combine -4xq and -8xq to get -12xq.

$$-3x^{2}-12xq+4q^{2}-4q^{2}$$

Combine 4q^{2} and -4q^{2} to get 0.

$$-3x^{2}-12xq$$

Expand

$$-3x^{2}-12qx$$

Solution Steps

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

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

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

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

Use the distributive property to multiply -4 by x^{2}+2xq+q^{2}.

$$x^{2}-4xq+4q^{2}-4x^{2}-8xq-4q^{2}$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}-4xq+4q^{2}-8xq-4q^{2}$$

Combine -4xq and -8xq to get -12xq.

$$-3x^{2}-12xq+4q^{2}-4q^{2}$$

Combine 4q^{2} and -4q^{2} to get 0.

$$-3x^{2}-12xq$$