Solve =x^2-14x+40

$$= x ^ { 2 } - 14 x + 40$$

Factor

$$\left(x-10\right)\left(x-4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 40=40$$

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 40.

$$-1,-40 -2,-20 -4,-10 -5,-8$$

Calculate the sum for each pair.

$$-1-40=-41 -2-20=-22 -4-10=-14 -5-8=-13$$

The solution is the pair that gives sum -14.

$$a=-10 b=-4$$

Rewrite x^{2}-14x+40 as \left(x^{2}-10x\right)+\left(-4x+40\right).

$$\left(x^{2}-10x\right)+\left(-4x+40\right)$$

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

$$x\left(x-10\right)-4\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x-4\right)$$

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.

$$x^{2}-14x+40=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 40}}{2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 40}}{2}$$

Multiply -4 times 40.

$$x=\frac{-\left(-14\right)±\sqrt{196-160}}{2}$$

Add 196 to -160.

$$x=\frac{-\left(-14\right)±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{-\left(-14\right)±6}{2}$$

The opposite of -14 is 14.

$$x=\frac{14±6}{2}$$

Now solve the equation x=\frac{14±6}{2} when ± is plus. Add 14 to 6.

$$x=\frac{20}{2}$$

Divide 20 by 2.

$$x=10$$

Now solve the equation x=\frac{14±6}{2} when ± is minus. Subtract 6 from 14.

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

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 10 for x_{1} and 4 for x_{2}.

$$x^{2}-14x+40=\left(x-10\right)\left(x-4\right)$$

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.

$$x ^ 2 -14x +40 = 0$$

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

$$r + s = 14 rs = 40$$

Two numbers r and s sum up to 14 exactly when the average of the two numbers is \frac{1}{2}*14 = 7. 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>

$$r = 7 - u s = 7 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 40

$$(7 - u) (7 + u) = 40$$

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

$$49 - u^2 = 40$$

Simplify the expression by subtracting 49 on both sides

$$-u^2 = 40-49 = -9$$

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.

$$r =7 - 3 = 4 s = 7 + 3 = 10$$

Evaluate

$$\left(x-10\right)\left(x-4\right)$$