Solve 4*(x^2+4x+4)-4*x*(x+5)<0

$$4 \times ( { x }^{ 2 } +4x+4)-4 \times x \times (x+5) < 0$$

Solve for x

$$x>4$$

Solution Steps

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

$$4x^{2}+16x+16-4x\left(x+5\right)<0$$

Use the distributive property to multiply -4x by x+5.

$$4x^{2}+16x+16-4x^{2}-20x<0$$

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

$$16x+16-20x<0$$

Combine 16x and -20x to get -4x.

$$-4x+16<0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-4x<-16$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x>\frac{-16}{-4}$$

Divide -16 by -4 to get 4.

$$x>4$$