Solve 3n^2+2n-33

$$3 n ^ { 2 } + 2 n - 33$$

Factor

$$\left(n-3\right)\left(3n+11\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3n^{2}+an+bn-33. To find a and b, set up a system to be solved.

$$a+b=2 ab=3\left(-33\right)=-99$$

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 -99.

$$-1,99 -3,33 -9,11$$

Calculate the sum for each pair.

$$-1+99=98 -3+33=30 -9+11=2$$

The solution is the pair that gives sum 2.

$$a=-9 b=11$$

Rewrite 3n^{2}+2n-33 as \left(3n^{2}-9n\right)+\left(11n-33\right).

$$\left(3n^{2}-9n\right)+\left(11n-33\right)$$

Factor out 3n in the first and 11 in the second group.

$$3n\left(n-3\right)+11\left(n-3\right)$$

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

$$\left(n-3\right)\left(3n+11\right)$$

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.

$$3n^{2}+2n-33=0$$

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.

$$n=\frac{-2±\sqrt{2^{2}-4\times 3\left(-33\right)}}{2\times 3}$$

Square 2.

$$n=\frac{-2±\sqrt{4-4\times 3\left(-33\right)}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-2±\sqrt{4-12\left(-33\right)}}{2\times 3}$$

Multiply -12 times -33.

$$n=\frac{-2±\sqrt{4+396}}{2\times 3}$$

Add 4 to 396.

$$n=\frac{-2±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$n=\frac{-2±20}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-2±20}{6}$$

Now solve the equation n=\frac{-2±20}{6} when ± is plus. Add -2 to 20.

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

Divide 18 by 6.

$$n=3$$

Now solve the equation n=\frac{-2±20}{6} when ± is minus. Subtract 20 from -2.

$$n=-\frac{22}{6}$$

Reduce the fraction \frac{-22}{6} to lowest terms by extracting and canceling out 2.

$$n=-\frac{11}{3}$$

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 -\frac{11}{3} for x_{2}.

$$3n^{2}+2n-33=3\left(n-3\right)\left(n-\left(-\frac{11}{3}\right)\right)$$

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

$$3n^{2}+2n-33=3\left(n-3\right)\left(n+\frac{11}{3}\right)$$

Add \frac{11}{3} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$3n^{2}+2n-33=3\left(n-3\right)\times \frac{3n+11}{3}$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3n^{2}+2n-33=\left(n-3\right)\left(3n+11\right)$$

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

$$x ^ 2 +\frac{2}{3}x -11 = 0$$

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

$$r + s = -\frac{2}{3} rs = -11$$

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

$$r = -\frac{1}{3} - u s = -\frac{1}{3} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -11

$$(-\frac{1}{3} - u) (-\frac{1}{3} + u) = -11$$

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

$$\frac{1}{9} - u^2 = -11$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -11-\frac{1}{9} = -\frac{100}{9}$$

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

$$u^2 = \frac{100}{9} u = \pm\sqrt{\frac{100}{9}} = \pm \frac{10}{3}$$

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

$$r =-\frac{1}{3} - \frac{10}{3} = -3.667 s = -\frac{1}{3} + \frac{10}{3} = 3$$

Evaluate

$$\left(n-3\right)\left(3n+11\right)$$