Solve 3x^2-15x-18

$$3 x ^ { 2 } - 15 x - 18$$

Factor

$$3\left(x-6\right)\left(x+1\right)$$

Solution Steps

Factor out 3.

$$3\left(x^{2}-5x-6\right)$$

Consider x^{2}-5x-6. Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-6\right)=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite x^{2}-5x-6 as \left(x^{2}-6x\right)+\left(x-6\right).

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

Factor out x in x^{2}-6x.

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

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

$$\left(x-6\right)\left(x+1\right)$$

Rewrite the complete factored expression.

$$3\left(x-6\right)\left(x+1\right)$$

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.

$$3x^{2}-15x-18=0$$

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.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\left(-18\right)}}{2\times 3}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 3\left(-18\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-15\right)±\sqrt{225-12\left(-18\right)}}{2\times 3}$$

Multiply -12 times -18.

$$x=\frac{-\left(-15\right)±\sqrt{225+216}}{2\times 3}$$

Add 225 to 216.

$$x=\frac{-\left(-15\right)±\sqrt{441}}{2\times 3}$$

Take the square root of 441.

$$x=\frac{-\left(-15\right)±21}{2\times 3}$$

The opposite of -15 is 15.

$$x=\frac{15±21}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{15±21}{6}$$

Now solve the equation x=\frac{15±21}{6} when ± is plus. Add 15 to 21.

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

Divide 36 by 6.

$$x=6$$

Now solve the equation x=\frac{15±21}{6} when ± is minus. Subtract 21 from 15.

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

Divide -6 by 6.

$$x=-1$$

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 -1 for x_{2}.

$$3x^{2}-15x-18=3\left(x-6\right)\left(x-\left(-1\right)\right)$$

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

$$3x^{2}-15x-18=3\left(x-6\right)\left(x+1\right)$$

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

$$x ^ 2 -5x -6 = 0$$

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

$$r + s = 5 rs = -6$$

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 = -6

$$(\frac{5}{2} - u) (\frac{5}{2} + u) = -6$$

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

$$\frac{25}{4} - u^2 = -6$$

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

$$-u^2 = -6-\frac{25}{4} = -\frac{49}{4}$$

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

$$u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}$$

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

$$r =\frac{5}{2} - \frac{7}{2} = -1 s = \frac{5}{2} + \frac{7}{2} = 6$$

Evaluate

$$3\left(x-6\right)\left(x+1\right)$$