Solve 3n^2+9n+6

$$3 n ^ { 2 } + 9 n + 6$$

Factor

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

Solution Steps

Factor out 3.

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

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

$$a+b=3 ab=1\times 2=2$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=1 b=2$$

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

$$\left(n^{2}+n\right)+\left(2n+2\right)$$

Factor out n in the first and 2 in the second group.

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

Factor out common term n+1 by using distributive property.

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

Rewrite the complete factored expression.

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

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.

$$3n^{2}+9n+6=0$$

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.

$$n=\frac{-9±\sqrt{9^{2}-4\times 3\times 6}}{2\times 3}$$

Square 9.

$$n=\frac{-9±\sqrt{81-4\times 3\times 6}}{2\times 3}$$

Multiply -4 times 3.

$$n=\frac{-9±\sqrt{81-12\times 6}}{2\times 3}$$

Multiply -12 times 6.

$$n=\frac{-9±\sqrt{81-72}}{2\times 3}$$

Add 81 to -72.

$$n=\frac{-9±\sqrt{9}}{2\times 3}$$

Take the square root of 9.

$$n=\frac{-9±3}{2\times 3}$$

Multiply 2 times 3.

$$n=\frac{-9±3}{6}$$

Now solve the equation n=\frac{-9±3}{6} when ± is plus. Add -9 to 3.

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

Divide -6 by 6.

$$n=-1$$

Now solve the equation n=\frac{-9±3}{6} when ± is minus. Subtract 3 from -9.

$$n=-\frac{12}{6}$$

Divide -12 by 6.

$$n=-2$$

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

$$3n^{2}+9n+6=3\left(n-\left(-1\right)\right)\left(n-\left(-2\right)\right)$$

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

$$3n^{2}+9n+6=3\left(n+1\right)\left(n+2\right)$$

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

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

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

$$r + s = -3 rs = 2$$

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-gzdabgg4ehffg0hf.b01.azurefd.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 = 2

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

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

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

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

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

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.

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

Evaluate

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