Solve 4x^2-20x+16

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

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}.

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

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 4

$$x ^ 2 -5x +4 = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = 5 rs = 4$$

Two numbers r and s sum up to 5 exactly when the average of the two numbers is \frac{1}{2}*5 = \frac{5}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{5}{2} - u s = \frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 4

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = 4$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{25}{4} - u^2 = 4$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = 4-\frac{25}{4} = -\frac{9}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{9}{4} u = \pm\sqrt{\frac{9}{4}} = \pm \frac{3}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{5}{2} - \frac{3}{2} = 1 s = \frac{5}{2} + \frac{3}{2} = 4$$

Evaluate

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