Solve 12x^2+20x+3=

$$12 x ^ { 2 } + 20 x + 3 =$$

Factor

$$\left(2x+3\right)\left(6x+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=20 ab=12\times 3=36$$

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 36.

$$1,36 2,18 3,12 4,9 6,6$$

Calculate the sum for each pair.

$$1+36=37 2+18=20 3+12=15 4+9=13 6+6=12$$

The solution is the pair that gives sum 20.

$$a=2 b=18$$

Rewrite 12x^{2}+20x+3 as \left(12x^{2}+2x\right)+\left(18x+3\right).

$$\left(12x^{2}+2x\right)+\left(18x+3\right)$$

Factor out 2x in the first and 3 in the second group.

$$2x\left(6x+1\right)+3\left(6x+1\right)$$

Factor out common term 6x+1 by using distributive property.

$$\left(6x+1\right)\left(2x+3\right)$$

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.

$$12x^{2}+20x+3=0$$

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.

$$x=\frac{-20±\sqrt{20^{2}-4\times 12\times 3}}{2\times 12}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 12\times 3}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-20±\sqrt{400-48\times 3}}{2\times 12}$$

Multiply -48 times 3.

$$x=\frac{-20±\sqrt{400-144}}{2\times 12}$$

Add 400 to -144.

$$x=\frac{-20±\sqrt{256}}{2\times 12}$$

Take the square root of 256.

$$x=\frac{-20±16}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-20±16}{24}$$

Now solve the equation x=\frac{-20±16}{24} when ± is plus. Add -20 to 16.

$$x=-\frac{4}{24}$$

Reduce the fraction \frac{-4}{24} to lowest terms by extracting and canceling out 4.

$$x=-\frac{1}{6}$$

Now solve the equation x=\frac{-20±16}{24} when ± is minus. Subtract 16 from -20.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{1}{6} for x_{1} and -\frac{3}{2} for x_{2}.

$$12x^{2}+20x+3=12\left(x-\left(-\frac{1}{6}\right)\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$12x^{2}+20x+3=12\left(x+\frac{1}{6}\right)\left(x+\frac{3}{2}\right)$$

Add \frac{1}{6} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+20x+3=12\times \frac{6x+1}{6}\left(x+\frac{3}{2}\right)$$

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

$$12x^{2}+20x+3=12\times \frac{6x+1}{6}\times \frac{2x+3}{2}$$

Multiply \frac{6x+1}{6} times \frac{2x+3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12x^{2}+20x+3=12\times \frac{\left(6x+1\right)\left(2x+3\right)}{6\times 2}$$

Multiply 6 times 2.

$$12x^{2}+20x+3=12\times \frac{\left(6x+1\right)\left(2x+3\right)}{12}$$

Cancel out 12, the greatest common factor in 12 and 12.

$$12x^{2}+20x+3=\left(6x+1\right)\left(2x+3\right)$$

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 12

$$x ^ 2 +\frac{5}{3}x +\frac{1}{4} = 0$$

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

$$r + s = -\frac{5}{3} rs = \frac{1}{4}$$

Two numbers r and s sum up to -\frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{3} = -\frac{5}{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>

$$r = -\frac{5}{6} - u s = -\frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{4}

$$(-\frac{5}{6} - u) (-\frac{5}{6} + u) = \frac{1}{4}$$

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

$$\frac{25}{36} - u^2 = \frac{1}{4}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = \frac{1}{4}-\frac{25}{36} = -\frac{4}{9}$$

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

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

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

$$r =-\frac{5}{6} - \frac{2}{3} = -1.500 s = -\frac{5}{6} + \frac{2}{3} = -0.167$$

Evaluate

$$\left(2x+3\right)\left(6x+1\right)$$