Solve 3t^2-12t-36

$$3 t ^ { 2 } - 12 t - 36$$

Factor

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

Solution Steps

Factor out 3.

$$3\left(t^{2}-4t-12\right)$$

Consider t^{2}-4t-12. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-12. To find a and b, set up a system to be solved.

$$a+b=-4 ab=1\left(-12\right)=-12$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -12.

$$1,-12 2,-6 3,-4$$

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -4.

$$a=-6 b=2$$

Rewrite t^{2}-4t-12 as \left(t^{2}-6t\right)+\left(2t-12\right).

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

Factor out t in the first and 2 in the second group.

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

Factor out common term t-6 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$3t^{2}-12t-36=0$$

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.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\left(-36\right)}}{2\times 3}$$

Square -12.

$$t=\frac{-\left(-12\right)±\sqrt{144-4\times 3\left(-36\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-\left(-12\right)±\sqrt{144-12\left(-36\right)}}{2\times 3}$$

Multiply -12 times -36.

$$t=\frac{-\left(-12\right)±\sqrt{144+432}}{2\times 3}$$

Add 144 to 432.

$$t=\frac{-\left(-12\right)±\sqrt{576}}{2\times 3}$$

Take the square root of 576.

$$t=\frac{-\left(-12\right)±24}{2\times 3}$$

The opposite of -12 is 12.

$$t=\frac{12±24}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{12±24}{6}$$

Now solve the equation t=\frac{12±24}{6} when ± is plus. Add 12 to 24.

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

Divide 36 by 6.

$$t=6$$

Now solve the equation t=\frac{12±24}{6} when ± is minus. Subtract 24 from 12.

$$t=-\frac{12}{6}$$

Divide -12 by 6.

$$t=-2$$

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

$$3t^{2}-12t-36=3\left(t-6\right)\left(t-\left(-2\right)\right)$$

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

$$3t^{2}-12t-36=3\left(t-6\right)\left(t+2\right)$$

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

$$x ^ 2 -4x -12 = 0$$

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

$$r + s = 4 rs = -12$$

Two numbers r and s sum up to 4 exactly when the average of the two numbers is \frac{1}{2}*4 = 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-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = 2 - u s = 2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -12

$$(2 - u) (2 + u) = -12$$

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

$$4 - u^2 = -12$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = -12-4 = -16$$

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

$$u^2 = 16 u = \pm\sqrt{16} = \pm 4$$

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

$$r =2 - 4 = -2 s = 2 + 4 = 6$$

Evaluate

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