Solve 3t^2+15t-252

$$3 t ^ { 2 } + 15 t - 252$$

Factor

$$3\left(t-7\right)\left(t+12\right)$$

Solution Steps

Factor out 3.

$$3\left(t^{2}+5t-84\right)$$

Consider t^{2}+5t-84. Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-84. To find a and b, set up a system to be solved.

$$a+b=5 ab=1\left(-84\right)=-84$$

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 -84.

$$-1,84 -2,42 -3,28 -4,21 -6,14 -7,12$$

Calculate the sum for each pair.

$$-1+84=83 -2+42=40 -3+28=25 -4+21=17 -6+14=8 -7+12=5$$

The solution is the pair that gives sum 5.

$$a=-7 b=12$$

Rewrite t^{2}+5t-84 as \left(t^{2}-7t\right)+\left(12t-84\right).

$$\left(t^{2}-7t\right)+\left(12t-84\right)$$

Factor out t in the first and 12 in the second group.

$$t\left(t-7\right)+12\left(t-7\right)$$

Factor out common term t-7 by using distributive property.

$$\left(t-7\right)\left(t+12\right)$$

Rewrite the complete factored expression.

$$3\left(t-7\right)\left(t+12\right)$$

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.

$$3t^{2}+15t-252=0$$

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.

$$t=\frac{-15±\sqrt{15^{2}-4\times 3\left(-252\right)}}{2\times 3}$$

Square 15.

$$t=\frac{-15±\sqrt{225-4\times 3\left(-252\right)}}{2\times 3}$$

Multiply -4 times 3.

$$t=\frac{-15±\sqrt{225-12\left(-252\right)}}{2\times 3}$$

Multiply -12 times -252.

$$t=\frac{-15±\sqrt{225+3024}}{2\times 3}$$

Add 225 to 3024.

$$t=\frac{-15±\sqrt{3249}}{2\times 3}$$

Take the square root of 3249.

$$t=\frac{-15±57}{2\times 3}$$

Multiply 2 times 3.

$$t=\frac{-15±57}{6}$$

Now solve the equation t=\frac{-15±57}{6} when ± is plus. Add -15 to 57.

$$t=\frac{42}{6}$$

Divide 42 by 6.

$$t=7$$

Now solve the equation t=\frac{-15±57}{6} when ± is minus. Subtract 57 from -15.

$$t=-\frac{72}{6}$$

Divide -72 by 6.

$$t=-12$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and -12 for x_{2}.

$$3t^{2}+15t-252=3\left(t-7\right)\left(t-\left(-12\right)\right)$$

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

$$3t^{2}+15t-252=3\left(t-7\right)\left(t+12\right)$$

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

$$x ^ 2 +5x -84 = 0$$

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

$$r + s = -5 rs = -84$$

Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{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>

$$r = -\frac{5}{2} - u s = -\frac{5}{2} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -84

$$(-\frac{5}{2} - u) (-\frac{5}{2} + u) = -84$$

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

$$\frac{25}{4} - u^2 = -84$$

Simplify the expression by subtracting \frac{25}{4} on both sides

$$-u^2 = -84-\frac{25}{4} = -\frac{361}{4}$$

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

$$u^2 = \frac{361}{4} u = \pm\sqrt{\frac{361}{4}} = \pm \frac{19}{2}$$

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

$$r =-\frac{5}{2} - \frac{19}{2} = -12 s = -\frac{5}{2} + \frac{19}{2} = 7$$

Evaluate

$$3\left(t-7\right)\left(t+12\right)$$