Solve 4x^2-20x+16

$$4 { x }^{ 2 } -20x+16$$

Factor

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

Solution Steps

Factor out 4.

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

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.

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

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.

$$4x^{2}-20x+16=0$$

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.

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

Square -20.

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

Multiply -4 times 4.

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

Multiply -16 times 16.

$$x=\frac{-\left(-20\right)±\sqrt{400-256}}{2\times 4}$$

Add 400 to -256.

$$x=\frac{-\left(-20\right)±\sqrt{144}}{2\times 4}$$

Take the square root of 144.

$$x=\frac{-\left(-20\right)±12}{2\times 4}$$

The opposite of -20 is 20.

$$x=\frac{20±12}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{20±12}{8}$$

Now solve the equation x=\frac{20±12}{8} when ± is plus. Add 20 to 12.

$$x=\frac{32}{8}$$

Divide 32 by 8.

$$x=4$$

Now solve the equation x=\frac{20±12}{8} when ± is minus. Subtract 12 from 20.

$$x=\frac{8}{8}$$

Divide 8 by 8.

$$x=1$$

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}.

$$4x^{2}-20x+16=4\left(x-4\right)\left(x-1\right)$$

Evaluate

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