Solve 3t^2+16t-12

$$3 t ^ { 2 } + 16 t - 12$$

Factor

$$\left(3t-2\right)\left(t+6\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3t^{2}+at+bt-12. To find a and b, set up a system to be solved.

$$a+b=16 ab=3\left(-12\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 16.

$$a=-2 b=18$$

Rewrite 3t^{2}+16t-12 as \left(3t^{2}-2t\right)+\left(18t-12\right).

$$\left(3t^{2}-2t\right)+\left(18t-12\right)$$

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

$$t\left(3t-2\right)+6\left(3t-2\right)$$

Factor out common term 3t-2 by using distributive property.

$$\left(3t-2\right)\left(t+6\right)$$

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.

$$3t^{2}+16t-12=0$$

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.

$$t=\frac{-16±\sqrt{16^{2}-4\times 3\left(-12\right)}}{2\times 3}$$

Square 16.

$$t=\frac{-16±\sqrt{256-4\times 3\left(-12\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-16±\sqrt{256-12\left(-12\right)}}{2\times 3}$$

Multiply -12 times -12.

$$t=\frac{-16±\sqrt{256+144}}{2\times 3}$$

Add 256 to 144.

$$t=\frac{-16±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$t=\frac{-16±20}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{-16±20}{6}$$

Now solve the equation t=\frac{-16±20}{6} when ± is plus. Add -16 to 20.

$$t=\frac{4}{6}$$

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$t=\frac{2}{3}$$

Now solve the equation t=\frac{-16±20}{6} when ± is minus. Subtract 20 from -16.

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

Divide -36 by 6.

$$t=-6$$

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

$$3t^{2}+16t-12=3\left(t-\frac{2}{3}\right)\left(t-\left(-6\right)\right)$$

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

$$3t^{2}+16t-12=3\left(t-\frac{2}{3}\right)\left(t+6\right)$$

Subtract \frac{2}{3} from t by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3t^{2}+16t-12=3\times \frac{3t-2}{3}\left(t+6\right)$$

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

$$3t^{2}+16t-12=\left(3t-2\right)\left(t+6\right)$$

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

$$x ^ 2 +\frac{16}{3}x -4 = 0$$

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

$$r + s = -\frac{16}{3} rs = -4$$

Two numbers r and s sum up to -\frac{16}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{16}{3} = -\frac{8}{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>

$$r = -\frac{8}{3} - u s = -\frac{8}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -4

$$(-\frac{8}{3} - u) (-\frac{8}{3} + u) = -4$$

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

$$\frac{64}{9} - u^2 = -4$$

Simplify the expression by subtracting \frac{64}{9} on both sides

$$-u^2 = -4-\frac{64}{9} = -\frac{100}{9}$$

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

$$u^2 = \frac{100}{9} u = \pm\sqrt{\frac{100}{9}} = \pm \frac{10}{3}$$

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

$$r =-\frac{8}{3} - \frac{10}{3} = -6 s = -\frac{8}{3} + \frac{10}{3} = 0.667$$

Evaluate

$$\left(3t-2\right)\left(t+6\right)$$