Solve x^2+25x+156

$$x ^ { 2 } + 25 x + 156$$

Factor

$$\left(x+12\right)\left(x+13\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+156. To find a and b, set up a system to be solved.

$$a+b=25 ab=1\times 156=156$$

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 156.

$$1,156 2,78 3,52 4,39 6,26 12,13$$

Calculate the sum for each pair.

$$1+156=157 2+78=80 3+52=55 4+39=43 6+26=32 12+13=25$$

The solution is the pair that gives sum 25.

$$a=12 b=13$$

Rewrite x^{2}+25x+156 as \left(x^{2}+12x\right)+\left(13x+156\right).

$$\left(x^{2}+12x\right)+\left(13x+156\right)$$

Factor out x in the first and 13 in the second group.

$$x\left(x+12\right)+13\left(x+12\right)$$

Factor out common term x+12 by using distributive property.

$$\left(x+12\right)\left(x+13\right)$$

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.

$$x^{2}+25x+156=0$$

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.

$$x=\frac{-25±\sqrt{25^{2}-4\times 156}}{2}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\times 156}}{2}$$

Multiply -4 times 156.

$$x=\frac{-25±\sqrt{625-624}}{2}$$

Add 625 to -624.

$$x=\frac{-25±\sqrt{1}}{2}$$

Take the square root of 1.

$$x=\frac{-25±1}{2}$$

Now solve the equation x=\frac{-25±1}{2} when ± is plus. Add -25 to 1.

$$x=-\frac{24}{2}$$

Divide -24 by 2.

$$x=-12$$

Now solve the equation x=\frac{-25±1}{2} when ± is minus. Subtract 1 from -25.

$$x=-\frac{26}{2}$$

Divide -26 by 2.

$$x=-13$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -12 for x_{1} and -13 for x_{2}.

$$x^{2}+25x+156=\left(x-\left(-12\right)\right)\left(x-\left(-13\right)\right)$$

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

$$x^{2}+25x+156=\left(x+12\right)\left(x+13\right)$$

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.

$$x ^ 2 +25x +156 = 0$$

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

$$r + s = -25 rs = 156$$

Two numbers r and s sum up to -25 exactly when the average of the two numbers is \frac{1}{2}*-25 = -\frac{25}{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>

$$r = -\frac{25}{2} - u s = -\frac{25}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 156

$$(-\frac{25}{2} - u) (-\frac{25}{2} + u) = 156$$

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

$$\frac{625}{4} - u^2 = 156$$

Simplify the expression by subtracting \frac{625}{4} on both sides

$$-u^2 = 156-\frac{625}{4} = -\frac{1}{4}$$

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.

$$r =-\frac{25}{2} - \frac{1}{2} = -13 s = -\frac{25}{2} + \frac{1}{2} = -12$$

Evaluate

$$\left(x+12\right)\left(x+13\right)$$