Solve x^2-16x+55

$$x ^ { 2 } - 16 x + 55$$

Factor

$$\left(x-11\right)\left(x-5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+55. To find a and b, set up a system to be solved.

$$a+b=-16 ab=1\times 55=55$$

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 55.

$$-1,-55 -5,-11$$

Calculate the sum for each pair.

$$-1-55=-56 -5-11=-16$$

The solution is the pair that gives sum -16.

$$a=-11 b=-5$$

Rewrite x^{2}-16x+55 as \left(x^{2}-11x\right)+\left(-5x+55\right).

$$\left(x^{2}-11x\right)+\left(-5x+55\right)$$

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

$$x\left(x-11\right)-5\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x-5\right)$$

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.

$$x^{2}-16x+55=0$$

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.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 55}}{2}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 55}}{2}$$

Multiply -4 times 55.

$$x=\frac{-\left(-16\right)±\sqrt{256-220}}{2}$$

Add 256 to -220.

$$x=\frac{-\left(-16\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-\left(-16\right)±6}{2}$$

The opposite of -16 is 16.

$$x=\frac{16±6}{2}$$

Now solve the equation x=\frac{16±6}{2} when ± is plus. Add 16 to 6.

$$x=\frac{22}{2}$$

Divide 22 by 2.

$$x=11$$

Now solve the equation x=\frac{16±6}{2} when ± is minus. Subtract 6 from 16.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 11 for x_{1} and 5 for x_{2}.

$$x^{2}-16x+55=\left(x-11\right)\left(x-5\right)$$

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.

$$x ^ 2 -16x +55 = 0$$

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

$$r + s = 16 rs = 55$$

Two numbers r and s sum up to 16 exactly when the average of the two numbers is \frac{1}{2}*16 = 8. 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>

$$r = 8 - u s = 8 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 55

$$(8 - u) (8 + u) = 55$$

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

$$64 - u^2 = 55$$

Simplify the expression by subtracting 64 on both sides

$$-u^2 = 55-64 = -9$$

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

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

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

$$r =8 - 3 = 5 s = 8 + 3 = 11$$

Evaluate

$$\left(x-11\right)\left(x-5\right)$$