Solve 1*(4x+1)(2x-2)>8x(x+45)

$$1 \cdot ( 4 x + 1 ) ( 2 x - 2 ) > 8 x ( x + 45 )$$

Solve for x

$$x<-\frac{1}{183}$$

Solution Steps

Use the distributive property to multiply 1 by 4x+1.

$$\left(4x+1\right)\left(2x-2\right)>8x\left(x+45\right)$$

Use the distributive property to multiply 4x+1 by 2x-2 and combine like terms.

$$8x^{2}-6x-2>8x\left(x+45\right)$$

Use the distributive property to multiply 8x by x+45.

$$8x^{2}-6x-2>8x^{2}+360x$$

Subtract 8x^{2} from both sides.

$$8x^{2}-6x-2-8x^{2}>360x$$

Combine 8x^{2} and -8x^{2} to get 0.

$$-6x-2>360x$$

Subtract 360x from both sides.

$$-6x-2-360x>0$$

Combine -6x and -360x to get -366x.

$$-366x-2>0$$

Add 2 to both sides. Anything plus zero gives itself.

$$-366x>2$$

Divide both sides by -366. Since -366 is negative, the inequality direction is changed.

$$x<\frac{2}{-366}$$

Reduce the fraction \frac{2}{-366} to lowest terms by extracting and canceling out 2.

$$x<-\frac{1}{183}$$