Solve 3x^2+2x-96

$$3 x ^ { 2 } + 2 x - 96$$

Factor

$$\left(3x-16\right)\left(x+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3x^{2}+ax+bx-96. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-96\right)=-288$$

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 -288.

$$-1,288 -2,144 -3,96 -4,72 -6,48 -8,36 -9,32 -12,24 -16,18$$

Calculate the sum for each pair.

$$-1+288=287 -2+144=142 -3+96=93 -4+72=68 -6+48=42 -8+36=28 -9+32=23 -12+24=12 -16+18=2$$

The solution is the pair that gives sum 2.

$$a=-16 b=18$$

Rewrite 3x^{2}+2x-96 as \left(3x^{2}-16x\right)+\left(18x-96\right).

$$\left(3x^{2}-16x\right)+\left(18x-96\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(3x-16\right)+6\left(3x-16\right)$$

Factor out common term 3x-16 by using distributive property.

$$\left(3x-16\right)\left(x+6\right)$$

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.

$$3x^{2}+2x-96=0$$

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.

$$x=\frac{-2±\sqrt{2^{2}-4\times 3\left(-96\right)}}{2\times 3}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 3\left(-96\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-2±\sqrt{4-12\left(-96\right)}}{2\times 3}$$

Multiply -12 times -96.

$$x=\frac{-2±\sqrt{4+1152}}{2\times 3}$$

Add 4 to 1152.

$$x=\frac{-2±\sqrt{1156}}{2\times 3}$$

Take the square root of 1156.

$$x=\frac{-2±34}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-2±34}{6}$$

Now solve the equation x=\frac{-2±34}{6} when ± is plus. Add -2 to 34.

$$x=\frac{32}{6}$$

Reduce the fraction \frac{32}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{16}{3}$$

Now solve the equation x=\frac{-2±34}{6} when ± is minus. Subtract 34 from -2.

$$x=-\frac{36}{6}$$

Divide -36 by 6.

$$x=-6$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{16}{3} for x_{1} and -6 for x_{2}.

$$3x^{2}+2x-96=3\left(x-\frac{16}{3}\right)\left(x-\left(-6\right)\right)$$

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

$$3x^{2}+2x-96=3\left(x-\frac{16}{3}\right)\left(x+6\right)$$

Subtract \frac{16}{3} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3x^{2}+2x-96=3\times \frac{3x-16}{3}\left(x+6\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3x^{2}+2x-96=\left(3x-16\right)\left(x+6\right)$$

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.This is achieved by dividing both sides of the equation by 3

$$x ^ 2 +\frac{2}{3}x -32 = 0$$

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

$$r + s = -\frac{2}{3} rs = -32$$

Two numbers r and s sum up to -\frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{2}{3} = -\frac{1}{3}. 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>

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -32

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -32$$

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

$$\frac{1}{9} - u^2 = -32$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -32-\frac{1}{9} = -\frac{289}{9}$$

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

$$u^2 = \frac{289}{9} u = \pm\sqrt{\frac{289}{9}} = \pm \frac{17}{3}$$

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

$$r =-\frac{1}{3} - \frac{17}{3} = -6 s = -\frac{1}{3} + \frac{17}{3} = 5.333$$

Evaluate

$$\left(3x-16\right)\left(x+6\right)$$