Solve 4(x-1)(x-2)<0

$$4 ( x - 1 ) ( x - 2 ) < 0$$

Solve for x

$$x\in \left(1,2\right)$$

Steps Using the Quadratic Formula

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

$$\left(4x-4\right)\left(x-2\right)<0$$

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

$$4x^{2}-12x+8<0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4x^{2}-12x+8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, -12 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\times 8}}{2\times 4}$$

Do the calculations.

$$x=\frac{12±4}{8}$$

Solve the equation x=\frac{12±4}{8} when ± is plus and when ± is minus.

$$x=2 x=1$$

Rewrite the inequality by using the obtained solutions.

$$4\left(x-2\right)\left(x-1\right)<0$$

For the product to be negative, x-2 and x-1 have to be of the opposite signs. Consider the case when x-2 is positive and x-1 is negative.

$$x-2>0 x-1<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-1 is positive and x-2 is negative.

$$x-1>0 x-2<0$$

The solution satisfying both inequalities is x\in \left(1,2\right).

$$x\in \left(1,2\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(1,2\right)$$