Solve 4x^2-20x+100

$$4 { x }^{ 2 } -20x+100$$

Factor

$$4\left(x^{2}-5x+25\right)$$

Solution Steps

Factor out 4. Polynomial x^{2}-5x+25 is not factored since it does not have any rational roots.

$$4\left(x^{2}-5x+25\right)$$

Steps Using the Quadratic Formula

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.

$$4x^{2}-20x+100=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 4\times 100}}{2\times 4}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 4\times 100}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-20\right)±\sqrt{400-16\times 100}}{2\times 4}$$

Multiply -16 times 100.

$$x=\frac{-\left(-20\right)±\sqrt{400-1600}}{2\times 4}$$

Add 400 to -1600.

$$x=\frac{-\left(-20\right)±\sqrt{-1200}}{2\times 4}$$

Since the square root of a negative number is not defined in the real field, there are no solutions. Quadratic polynomial cannot be factored.

$$4x^{2}-20x+100$$

Evaluate

$$4\left(x^{2}-5x+25\right)$$