Solve -4x^2+20x-25

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

Factor

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

Solution Steps

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

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

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

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

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

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

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

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

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.

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

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.

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

Square 20.

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

Multiply -4 times -4.

$$x=\frac{-20±\sqrt{400+16\left(-25\right)}}{2\left(-4\right)}$$

Multiply 16 times -25.

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

Add 400 to -400.

$$x=\frac{-20±\sqrt{0}}{2\left(-4\right)}$$

Take the square root of 0.

$$x=\frac{-20±0}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-20±0}{-8}$$

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

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

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

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

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

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

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

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

Multiply -2 times -2.

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

Cancel out 4, the greatest common factor in -4 and 4.

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

Evaluate

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