Solve 9n^2+36n+20

$$9 n ^ { 2 } + 36 n + 20$$

Factor

$$\left(3n+2\right)\left(3n+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 9n^{2}+an+bn+20. To find a and b, set up a system to be solved.

$$a+b=36 ab=9\times 20=180$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 180.

$$1,180 2,90 3,60 4,45 5,36 6,30 9,20 10,18 12,15$$

Calculate the sum for each pair.

$$1+180=181 2+90=92 3+60=63 4+45=49 5+36=41 6+30=36 9+20=29 10+18=28 12+15=27$$

The solution is the pair that gives sum 36.

$$a=6 b=30$$

Rewrite 9n^{2}+36n+20 as \left(9n^{2}+6n\right)+\left(30n+20\right).

$$\left(9n^{2}+6n\right)+\left(30n+20\right)$$

Factor out 3n in the first and 10 in the second group.

$$3n\left(3n+2\right)+10\left(3n+2\right)$$

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

$$\left(3n+2\right)\left(3n+10\right)$$

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.

$$9n^{2}+36n+20=0$$

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.

$$n=\frac{-36±\sqrt{36^{2}-4\times 9\times 20}}{2\times 9}$$

Square 36.

$$n=\frac{-36±\sqrt{1296-4\times 9\times 20}}{2\times 9}$$

Multiply -4 times 9.

$$n=\frac{-36±\sqrt{1296-36\times 20}}{2\times 9}$$

Multiply -36 times 20.

$$n=\frac{-36±\sqrt{1296-720}}{2\times 9}$$

Add 1296 to -720.

$$n=\frac{-36±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$n=\frac{-36±24}{2\times 9}$$

Multiply 2 times 9.

$$n=\frac{-36±24}{18}$$

Now solve the equation n=\frac{-36±24}{18} when ± is plus. Add -36 to 24.

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

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{-36±24}{18} when ± is minus. Subtract 24 from -36.

$$n=-\frac{60}{18}$$

Reduce the fraction \frac{-60}{18} to lowest terms by extracting and canceling out 6.

$$n=-\frac{10}{3}$$

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{10}{3} for x_{2}.

$$9n^{2}+36n+20=9\left(n-\left(-\frac{2}{3}\right)\right)\left(n-\left(-\frac{10}{3}\right)\right)$$

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

$$9n^{2}+36n+20=9\left(n+\frac{2}{3}\right)\left(n+\frac{10}{3}\right)$$

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

$$9n^{2}+36n+20=9\times \frac{3n+2}{3}\left(n+\frac{10}{3}\right)$$

Add \frac{10}{3} to n by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$9n^{2}+36n+20=9\times \frac{3n+2}{3}\times \frac{3n+10}{3}$$

Multiply \frac{3n+2}{3} times \frac{3n+10}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$9n^{2}+36n+20=9\times \frac{\left(3n+2\right)\left(3n+10\right)}{3\times 3}$$

Multiply 3 times 3.

$$9n^{2}+36n+20=9\times \frac{\left(3n+2\right)\left(3n+10\right)}{9}$$

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

$$9n^{2}+36n+20=\left(3n+2\right)\left(3n+10\right)$$

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

$$x ^ 2 +4x +\frac{20}{9} = 0$$

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

$$r + s = -4 rs = \frac{20}{9}$$

Two numbers r and s sum up to -4 exactly when the average of the two numbers is \frac{1}{2}*-4 = -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>

$$r = -2 - u s = -2 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{20}{9}

$$(-2 - u) (-2 + u) = \frac{20}{9}$$

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

$$4 - u^2 = \frac{20}{9}$$

Simplify the expression by subtracting 4 on both sides

$$-u^2 = \frac{20}{9}-4 = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =-2 - \frac{4}{3} = -3.333 s = -2 + \frac{4}{3} = -0.667$$

Evaluate

$$\left(3n+2\right)\left(3n+10\right)$$