Solve 56-18x+x^2

$$56-18x+ { x }^{ 2 }$$

Factor

$$\left(x-14\right)\left(x-4\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$x^{2}-18x+56$$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+56. To find a and b, set up a system to be solved.

$$a+b=-18 ab=1\times 56=56$$

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 56.

$$-1,-56 -2,-28 -4,-14 -7,-8$$

Calculate the sum for each pair.

$$-1-56=-57 -2-28=-30 -4-14=-18 -7-8=-15$$

The solution is the pair that gives sum -18.

$$a=-14 b=-4$$

Rewrite x^{2}-18x+56 as \left(x^{2}-14x\right)+\left(-4x+56\right).

$$\left(x^{2}-14x\right)+\left(-4x+56\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-14\right)-4\left(x-14\right)$$

Factor out common term x-14 by using distributive property.

$$\left(x-14\right)\left(x-4\right)$$

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.

$$x^{2}-18x+56=0$$

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.

$$x=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 56}}{2}$$

Square -18.

$$x=\frac{-\left(-18\right)±\sqrt{324-4\times 56}}{2}$$

Multiply -4 times 56.

$$x=\frac{-\left(-18\right)±\sqrt{324-224}}{2}$$

Add 324 to -224.

$$x=\frac{-\left(-18\right)±\sqrt{100}}{2}$$

Take the square root of 100.

$$x=\frac{-\left(-18\right)±10}{2}$$

The opposite of -18 is 18.

$$x=\frac{18±10}{2}$$

Now solve the equation x=\frac{18±10}{2} when ± is plus. Add 18 to 10.

$$x=\frac{28}{2}$$

Divide 28 by 2.

$$x=14$$

Now solve the equation x=\frac{18±10}{2} when ± is minus. Subtract 10 from 18.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 14 for x_{1} and 4 for x_{2}.

$$x^{2}-18x+56=\left(x-14\right)\left(x-4\right)$$

Evaluate

$$\left(x-14\right)\left(x-4\right)$$