Solve 18x^2+36x+16

$$18 x ^ { 2 } + 36 x + 16$$

Factor

$$2\left(3x+2\right)\left(3x+4\right)$$

Solution Steps

Factor out 2.

$$2\left(9x^{2}+18x+8\right)$$

Consider 9x^{2}+18x+8. Factor the expression by grouping. First, the expression needs to be rewritten as 9x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=18 ab=9\times 8=72$$

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 72.

$$1,72 2,36 3,24 4,18 6,12 8,9$$

Calculate the sum for each pair.

$$1+72=73 2+36=38 3+24=27 4+18=22 6+12=18 8+9=17$$

The solution is the pair that gives sum 18.

$$a=6 b=12$$

Rewrite 9x^{2}+18x+8 as \left(9x^{2}+6x\right)+\left(12x+8\right).

$$\left(9x^{2}+6x\right)+\left(12x+8\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(3x+2\right)+4\left(3x+2\right)$$

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

$$\left(3x+2\right)\left(3x+4\right)$$

Rewrite the complete factored expression.

$$2\left(3x+2\right)\left(3x+4\right)$$

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.

$$18x^{2}+36x+16=0$$

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.

$$x=\frac{-36±\sqrt{36^{2}-4\times 18\times 16}}{2\times 18}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\times 18\times 16}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-36±\sqrt{1296-72\times 16}}{2\times 18}$$

Multiply -72 times 16.

$$x=\frac{-36±\sqrt{1296-1152}}{2\times 18}$$

Add 1296 to -1152.

$$x=\frac{-36±\sqrt{144}}{2\times 18}$$

Take the square root of 144.

$$x=\frac{-36±12}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-36±12}{36}$$

Now solve the equation x=\frac{-36±12}{36} when ± is plus. Add -36 to 12.

$$x=-\frac{24}{36}$$

Reduce the fraction \frac{-24}{36} to lowest terms by extracting and canceling out 12.

$$x=-\frac{2}{3}$$

Now solve the equation x=\frac{-36±12}{36} when ± is minus. Subtract 12 from -36.

$$x=-\frac{48}{36}$$

Reduce the fraction \frac{-48}{36} to lowest terms by extracting and canceling out 12.

$$x=-\frac{4}{3}$$

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{4}{3} for x_{2}.

$$18x^{2}+36x+16=18\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-\frac{4}{3}\right)\right)$$

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

$$18x^{2}+36x+16=18\left(x+\frac{2}{3}\right)\left(x+\frac{4}{3}\right)$$

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

$$18x^{2}+36x+16=18\times \frac{3x+2}{3}\left(x+\frac{4}{3}\right)$$

Add \frac{4}{3} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+36x+16=18\times \frac{3x+2}{3}\times \frac{3x+4}{3}$$

Multiply \frac{3x+2}{3} times \frac{3x+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$18x^{2}+36x+16=18\times \frac{\left(3x+2\right)\left(3x+4\right)}{3\times 3}$$

Multiply 3 times 3.

$$18x^{2}+36x+16=18\times \frac{\left(3x+2\right)\left(3x+4\right)}{9}$$

Cancel out 9, the greatest common factor in 18 and 9.

$$18x^{2}+36x+16=2\left(3x+2\right)\left(3x+4\right)$$

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 18

$$x ^ 2 +2x +\frac{8}{9} = 0$$

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

$$r + s = -2 rs = \frac{8}{9}$$

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

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{8}{9}

$$(-1 - u) (-1 + u) = \frac{8}{9}$$

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

$$1 - u^2 = \frac{8}{9}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = \frac{8}{9}-1 = -\frac{1}{9}$$

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

$$u^2 = \frac{1}{9} u = \pm\sqrt{\frac{1}{9}} = \pm \frac{1}{3}$$

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

$$r =-1 - \frac{1}{3} = -1.333 s = -1 + \frac{1}{3} = -0.667$$

Evaluate

$$18x^{2}+36x+16$$