Solve 5x^2-25x+20

$$5 { x }^{ 2 } -25x+20$$

Factor

$$5\left(x-4\right)\left(x-1\right)$$

Solution Steps

Factor out 5.

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

Consider x^{2}-5x+4. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\times 4=4$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -5.

$$a=-4 b=-1$$

Rewrite x^{2}-5x+4 as \left(x^{2}-4x\right)+\left(-x+4\right).

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

Factor out x in the first and -1 in the second group.

$$x\left(x-4\right)-\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x-1\right)$$

Rewrite the complete factored expression.

$$5\left(x-4\right)\left(x-1\right)$$

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.

$$5x^{2}-25x+20=0$$

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.

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

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 5\times 20}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-25\right)±\sqrt{625-20\times 20}}{2\times 5}$$

Multiply -20 times 20.

$$x=\frac{-\left(-25\right)±\sqrt{625-400}}{2\times 5}$$

Add 625 to -400.

$$x=\frac{-\left(-25\right)±\sqrt{225}}{2\times 5}$$

Take the square root of 225.

$$x=\frac{-\left(-25\right)±15}{2\times 5}$$

The opposite of -25 is 25.

$$x=\frac{25±15}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{25±15}{10}$$

Now solve the equation x=\frac{25±15}{10} when ± is plus. Add 25 to 15.

$$x=\frac{40}{10}$$

Divide 40 by 10.

$$x=4$$

Now solve the equation x=\frac{25±15}{10} when ± is minus. Subtract 15 from 25.

$$x=\frac{10}{10}$$

Divide 10 by 10.

$$x=1$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and 1 for x_{2}.

$$5x^{2}-25x+20=5\left(x-4\right)\left(x-1\right)$$

Evaluate

$$5\left(x-4\right)\left(x-1\right)$$