Solve 9n^2-3n-2?

$$9 n ^ { 2 } - 3 n - 2 ?$$

Factor

$$\left(3n-2\right)\left(3n+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9n^{2}+an+bn-2. To find a and b, set up a system to be solved.

$$a+b=-3 ab=9\left(-2\right)=-18$$

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 -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.

$$a=-6 b=3$$

Rewrite 9n^{2}-3n-2 as \left(9n^{2}-6n\right)+\left(3n-2\right).

$$\left(9n^{2}-6n\right)+\left(3n-2\right)$$

Factor out 3n in 9n^{2}-6n.

$$3n\left(3n-2\right)+3n-2$$

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

$$\left(3n-2\right)\left(3n+1\right)$$

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.

$$9n^{2}-3n-2=0$$

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.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 9\left(-2\right)}}{2\times 9}$$

Square -3.

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

Multiply -4 times 9.

$$n=\frac{-\left(-3\right)±\sqrt{9-36\left(-2\right)}}{2\times 9}$$

Multiply -36 times -2.

$$n=\frac{-\left(-3\right)±\sqrt{9+72}}{2\times 9}$$

Add 9 to 72.

$$n=\frac{-\left(-3\right)±\sqrt{81}}{2\times 9}$$

Take the square root of 81.

$$n=\frac{-\left(-3\right)±9}{2\times 9}$$

The opposite of -3 is 3.

$$n=\frac{3±9}{2\times 9}$$

Multiply 2 times 9.

$$n=\frac{3±9}{18}$$

Now solve the equation n=\frac{3±9}{18} when ± is plus. Add 3 to 9.

$$n=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

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

Now solve the equation n=\frac{3±9}{18} when ± is minus. Subtract 9 from 3.

$$n=-\frac{6}{18}$$

Reduce the fraction \frac{-6}{18} to lowest terms by extracting and canceling out 6.

$$n=-\frac{1}{3}$$

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 -\frac{1}{3} for x_{2}.

$$9n^{2}-3n-2=9\left(n-\frac{2}{3}\right)\left(n-\left(-\frac{1}{3}\right)\right)$$

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

$$9n^{2}-3n-2=9\left(n-\frac{2}{3}\right)\left(n+\frac{1}{3}\right)$$

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

$$9n^{2}-3n-2=9\times \frac{3n-2}{3}\left(n+\frac{1}{3}\right)$$

Add \frac{1}{3} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9n^{2}-3n-2=9\times \frac{3n-2}{3}\times \frac{3n+1}{3}$$

Multiply \frac{3n-2}{3} times \frac{3n+1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9n^{2}-3n-2=9\times \frac{\left(3n-2\right)\left(3n+1\right)}{3\times 3}$$

Multiply 3 times 3.

$$9n^{2}-3n-2=9\times \frac{\left(3n-2\right)\left(3n+1\right)}{9}$$

Cancel out 9, the greatest common factor in 9 and 9.

$$9n^{2}-3n-2=\left(3n-2\right)\left(3n+1\right)$$

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 9

$$x ^ 2 -\frac{1}{3}x -\frac{2}{9} = 0$$

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

$$r + s = \frac{1}{3} rs = -\frac{2}{9}$$

Two numbers r and s sum up to \frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{3} = \frac{1}{6}. 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>

$$r = \frac{1}{6} - u s = \frac{1}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{9}

$$(\frac{1}{6} - u) (\frac{1}{6} + u) = -\frac{2}{9}$$

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

$$\frac{1}{36} - u^2 = -\frac{2}{9}$$

Simplify the expression by subtracting \frac{1}{36} on both sides

$$-u^2 = -\frac{2}{9}-\frac{1}{36} = -\frac{1}{4}$$

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

$$u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{2}$$

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

$$r =\frac{1}{6} - \frac{1}{2} = -0.333 s = \frac{1}{6} + \frac{1}{2} = 0.667$$

Evaluate

$$\left(3n-2\right)\left(3n+1\right)$$