Solve x^2+15x+56

$$x ^ { 2 } + 15 x + 56$$

Factor

$$\left(x+7\right)\left(x+8\right)$$

Solution Steps

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.

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

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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.

$$a=7 b=8$$

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

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

Factor out x in the first and 8 in the second group.

$$x\left(x+7\right)+8\left(x+7\right)$$

Factor out common term x+7 by using distributive property.

$$\left(x+7\right)\left(x+8\right)$$

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.

$$x^{2}+15x+56=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\times 56}}{2}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 56}}{2}$$

Multiply -4 times 56.

$$x=\frac{-15±\sqrt{225-224}}{2}$$

Add 225 to -224.

$$x=\frac{-15±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-15±1}{2}$$

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

$$x=-\frac{14}{2}$$

Divide -14 by 2.

$$x=-7$$

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

$$x=-\frac{16}{2}$$

Divide -16 by 2.

$$x=-8$$

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

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

Simplify all the expressions of the form p-\left(-q\right) to p+q.

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

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.

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

Evaluate

$$\left(x+7\right)\left(x+8\right)$$