Solve t^2+3t-18

$$t ^ { 2 } + 3 t - 18$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as t^{2}+at+bt-18. To find a and b, set up a system to be solved.

$$a+b=3 ab=1\left(-18\right)=-18$$

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 -18.

$$-1,18 -2,9 -3,6$$

Calculate the sum for each pair.

$$-1+18=17 -2+9=7 -3+6=3$$

The solution is the pair that gives sum 3.

$$a=-3 b=6$$

Rewrite t^{2}+3t-18 as \left(t^{2}-3t\right)+\left(6t-18\right).

$$\left(t^{2}-3t\right)+\left(6t-18\right)$$

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

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

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

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

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.

$$t^{2}+3t-18=0$$

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.

$$t=\frac{-3±\sqrt{3^{2}-4\left(-18\right)}}{2}$$

Square 3.

$$t=\frac{-3±\sqrt{9-4\left(-18\right)}}{2}$$

Multiply -4 times -18.

$$t=\frac{-3±\sqrt{9+72}}{2}$$

Add 9 to 72.

$$t=\frac{-3±\sqrt{81}}{2}$$

Take the square root of 81.

$$t=\frac{-3±9}{2}$$

Now solve the equation t=\frac{-3±9}{2} when ± is plus. Add -3 to 9.

$$t=\frac{6}{2}$$

Divide 6 by 2.

$$t=3$$

Now solve the equation t=\frac{-3±9}{2} when ± is minus. Subtract 9 from -3.

$$t=-\frac{12}{2}$$

Divide -12 by 2.

$$t=-6$$

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

$$t^{2}+3t-18=\left(t-3\right)\left(t-\left(-6\right)\right)$$

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

$$t^{2}+3t-18=\left(t-3\right)\left(t+6\right)$$

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.

$$x ^ 2 +3x -18 = 0$$

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

$$r + s = -3 rs = -18$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -18

$$(-\frac{3}{2} - u) (-\frac{3}{2} + u) = -18$$

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

$$\frac{9}{4} - u^2 = -18$$

Simplify the expression by subtracting \frac{9}{4} on both sides

$$-u^2 = -18-\frac{9}{4} = -\frac{81}{4}$$

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

$$u^2 = \frac{81}{4} u = \pm\sqrt{\frac{81}{4}} = \pm \frac{9}{2}$$

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

$$r =-\frac{3}{2} - \frac{9}{2} = -6 s = -\frac{3}{2} + \frac{9}{2} = 3$$

Evaluate

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