Solve 12x^2+17x+6

$$12 x ^ { 2 } + 17 x + 6$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=17 ab=12\times 6=72$$

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 17.

$$a=8 b=9$$

Rewrite 12x^{2}+17x+6 as \left(12x^{2}+8x\right)+\left(9x+6\right).

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

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

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

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

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

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.

$$12x^{2}+17x+6=0$$

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.

$$x=\frac{-17±\sqrt{17^{2}-4\times 12\times 6}}{2\times 12}$$

Square 17.

$$x=\frac{-17±\sqrt{289-4\times 12\times 6}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-17±\sqrt{289-48\times 6}}{2\times 12}$$

Multiply -48 times 6.

$$x=\frac{-17±\sqrt{289-288}}{2\times 12}$$

Add 289 to -288.

$$x=\frac{-17±\sqrt{1}}{2\times 12}$$

Take the square root of 1.

$$x=\frac{-17±1}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-17±1}{24}$$

Now solve the equation x=\frac{-17±1}{24} when ± is plus. Add -17 to 1.

$$x=-\frac{16}{24}$$

Reduce the fraction \frac{-16}{24} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{-17±1}{24} when ± is minus. Subtract 1 from -17.

$$x=-\frac{18}{24}$$

Reduce the fraction \frac{-18}{24} to lowest terms by extracting and canceling out 6.

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

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

$$12x^{2}+17x+6=12\left(x-\left(-\frac{2}{3}\right)\right)\left(x-\left(-\frac{3}{4}\right)\right)$$

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

$$12x^{2}+17x+6=12\left(x+\frac{2}{3}\right)\left(x+\frac{3}{4}\right)$$

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

$$12x^{2}+17x+6=12\times \frac{3x+2}{3}\left(x+\frac{3}{4}\right)$$

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

$$12x^{2}+17x+6=12\times \frac{3x+2}{3}\times \frac{4x+3}{4}$$

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

$$12x^{2}+17x+6=12\times \frac{\left(3x+2\right)\left(4x+3\right)}{3\times 4}$$

Multiply 3 times 4.

$$12x^{2}+17x+6=12\times \frac{\left(3x+2\right)\left(4x+3\right)}{12}$$

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

$$12x^{2}+17x+6=\left(3x+2\right)\left(4x+3\right)$$

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

$$x ^ 2 +\frac{17}{12}x +\frac{1}{2} = 0$$

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

$$r + s = -\frac{17}{12} rs = \frac{1}{2}$$

Two numbers r and s sum up to -\frac{17}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{12} = -\frac{17}{24}. 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>

$$r = -\frac{17}{24} - u s = -\frac{17}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}

$$(-\frac{17}{24} - u) (-\frac{17}{24} + u) = \frac{1}{2}$$

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

$$\frac{289}{576} - u^2 = \frac{1}{2}$$

Simplify the expression by subtracting \frac{289}{576} on both sides

$$-u^2 = \frac{1}{2}-\frac{289}{576} = -\frac{1}{576}$$

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

$$u^2 = \frac{1}{576} u = \pm\sqrt{\frac{1}{576}} = \pm \frac{1}{24}$$

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

$$r =-\frac{17}{24} - \frac{1}{24} = -0.750 s = -\frac{17}{24} + \frac{1}{24} = -0.667$$

Evaluate

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