Solve x^2+x-306

$$x ^ { 2 } + x - 306$$

Factor

$$\left(x-17\right)\left(x+18\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-306. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-306\right)=-306$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -306.

$$-1,306 -2,153 -3,102 -6,51 -9,34 -17,18$$

Calculate the sum for each pair.

$$-1+306=305 -2+153=151 -3+102=99 -6+51=45 -9+34=25 -17+18=1$$

The solution is the pair that gives sum 1.

$$a=-17 b=18$$

Rewrite x^{2}+x-306 as \left(x^{2}-17x\right)+\left(18x-306\right).

$$\left(x^{2}-17x\right)+\left(18x-306\right)$$

Factor out x in the first and 18 in the second group.

$$x\left(x-17\right)+18\left(x-17\right)$$

Factor out common term x-17 by using distributive property.

$$\left(x-17\right)\left(x+18\right)$$

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.

$$x^{2}+x-306=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-306\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-306\right)}}{2}$$

Multiply -4 times -306.

$$x=\frac{-1±\sqrt{1+1224}}{2}$$

Add 1 to 1224.

$$x=\frac{-1±\sqrt{1225}}{2}$$

Take the square root of 1225.

$$x=\frac{-1±35}{2}$$

Now solve the equation x=\frac{-1±35}{2} when ± is plus. Add -1 to 35.

$$x=\frac{34}{2}$$

Divide 34 by 2.

$$x=17$$

Now solve the equation x=\frac{-1±35}{2} when ± is minus. Subtract 35 from -1.

$$x=-\frac{36}{2}$$

Divide -36 by 2.

$$x=-18$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 17 for x_{1} and -18 for x_{2}.

$$x^{2}+x-306=\left(x-17\right)\left(x-\left(-18\right)\right)$$

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

$$x^{2}+x-306=\left(x-17\right)\left(x+18\right)$$

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.

$$x ^ 2 +1x -306 = 0$$

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

$$r + s = -1 rs = -306$$

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

$$r = -\frac{1}{2} - u s = -\frac{1}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -306

$$(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -306$$

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

$$\frac{1}{4} - u^2 = -306$$

Simplify the expression by subtracting \frac{1}{4} on both sides

$$-u^2 = -306-\frac{1}{4} = -\frac{1225}{4}$$

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

$$u^2 = \frac{1225}{4} u = \pm\sqrt{\frac{1225}{4}} = \pm \frac{35}{2}$$

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

$$r =-\frac{1}{2} - \frac{35}{2} = -18 s = -\frac{1}{2} + \frac{35}{2} = 17$$

Evaluate

$$\left(x-17\right)\left(x+18\right)$$