Solve c^2-15c+56

$$c ^ { 2 } - 15 c + 56$$

Factor

$$\left(c-8\right)\left(c-7\right)$$

Solution Steps

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

$$a+b=-15 ab=1\times 56=56$$

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 -15.

$$a=-8 b=-7$$

Rewrite c^{2}-15c+56 as \left(c^{2}-8c\right)+\left(-7c+56\right).

$$\left(c^{2}-8c\right)+\left(-7c+56\right)$$

Factor out c in the first and -7 in the second group.

$$c\left(c-8\right)-7\left(c-8\right)$$

Factor out common term c-8 by using distributive property.

$$\left(c-8\right)\left(c-7\right)$$

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.

$$c^{2}-15c+56=0$$

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.

$$c=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 56}}{2}$$

Square -15.

$$c=\frac{-\left(-15\right)±\sqrt{225-4\times 56}}{2}$$

Multiply -4 times 56.

$$c=\frac{-\left(-15\right)±\sqrt{225-224}}{2}$$

Add 225 to -224.

$$c=\frac{-\left(-15\right)±\sqrt{1}}{2}$$

Take the square root of 1.

$$c=\frac{-\left(-15\right)±1}{2}$$

The opposite of -15 is 15.

$$c=\frac{15±1}{2}$$

Now solve the equation c=\frac{15±1}{2} when ± is plus. Add 15 to 1.

$$c=\frac{16}{2}$$

Divide 16 by 2.

$$c=8$$

Now solve the equation c=\frac{15±1}{2} when ± is minus. Subtract 1 from 15.

$$c=\frac{14}{2}$$

Divide 14 by 2.

$$c=7$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 8 for x_{1} and 7 for x_{2}.

$$c^{2}-15c+56=\left(c-8\right)\left(c-7\right)$$

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.

$$x ^ 2 -15x +56 = 0$$

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

$$r + s = 15 rs = 56$$

Two numbers r and s sum up to 15 exactly when the average of the two numbers is \frac{1}{2}*15 = \frac{15}{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>

$$r = \frac{15}{2} - u s = \frac{15}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 56

$$(\frac{15}{2} - u) (\frac{15}{2} + u) = 56$$

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

$$\frac{225}{4} - u^2 = 56$$

Simplify the expression by subtracting \frac{225}{4} on both sides

$$-u^2 = 56-\frac{225}{4} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =\frac{15}{2} - \frac{1}{2} = 7 s = \frac{15}{2} + \frac{1}{2} = 8$$

Evaluate

$$\left(c-8\right)\left(c-7\right)$$