Solve -4+20x-25x^2

$$-4+20x-25 { x }^{ 2 }$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-25x^{2}+20x-4$$

Factor the expression by grouping. First, the expression needs to be rewritten as -25x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=20 ab=-25\left(-4\right)=100$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 100.

$$1,100 2,50 4,25 5,20 10,10$$

Calculate the sum for each pair.

$$1+100=101 2+50=52 4+25=29 5+20=25 10+10=20$$

The solution is the pair that gives sum 20.

$$a=10 b=10$$

Rewrite -25x^{2}+20x-4 as \left(-25x^{2}+10x\right)+\left(10x-4\right).

$$\left(-25x^{2}+10x\right)+\left(10x-4\right)$$

Factor out -5x in the first and 2 in the second group.

$$-5x\left(5x-2\right)+2\left(5x-2\right)$$

Factor out common term 5x-2 by using distributive property.

$$\left(5x-2\right)\left(-5x+2\right)$$

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.

$$-25x^{2}+20x-4=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-25\right)\left(-4\right)}}{2\left(-25\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-25\right)\left(-4\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-20±\sqrt{400+100\left(-4\right)}}{2\left(-25\right)}$$

Multiply 100 times -4.

$$x=\frac{-20±\sqrt{400-400}}{2\left(-25\right)}$$

Add 400 to -400.

$$x=\frac{-20±\sqrt{0}}{2\left(-25\right)}$$

Take the square root of 0.

$$x=\frac{-20±0}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-20±0}{-50}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{2}{5} for x_{1} and \frac{2}{5} for x_{2}.

$$-25x^{2}+20x-4=-25\left(x-\frac{2}{5}\right)\left(x-\frac{2}{5}\right)$$

Subtract \frac{2}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-25x^{2}+20x-4=-25\times \frac{-5x+2}{-5}\left(x-\frac{2}{5}\right)$$

Subtract \frac{2}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-25x^{2}+20x-4=-25\times \frac{-5x+2}{-5}\times \frac{-5x+2}{-5}$$

Multiply \frac{-5x+2}{-5} times \frac{-5x+2}{-5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$-25x^{2}+20x-4=-25\times \frac{\left(-5x+2\right)\left(-5x+2\right)}{-5\left(-5\right)}$$

Multiply -5 times -5.

$$-25x^{2}+20x-4=-25\times \frac{\left(-5x+2\right)\left(-5x+2\right)}{25}$$

Cancel out 25, the greatest common factor in -25 and 25.

$$-25x^{2}+20x-4=-\left(-5x+2\right)\left(-5x+2\right)$$

Evaluate

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