Solve f(x)=36x^2+15x-6

$$f ( x ) = 36 x ^ { 2 } + 15 x - 6$$

Factor

$$3\left(4x-1\right)\left(3x+2\right)$$

Solution Steps

Factor out 3.

$$3\left(12x^{2}+5x-2\right)$$

Consider 12x^{2}+5x-2. Factor the expression by grouping. First, the expression needs to be rewritten as 12x^{2}+ax+bx-2. To find a and b, set up a system to be solved.

$$a+b=5 ab=12\left(-2\right)=-24$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -24.

$$-1,24 -2,12 -3,8 -4,6$$

Calculate the sum for each pair.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

The solution is the pair that gives sum 5.

$$a=-3 b=8$$

Rewrite 12x^{2}+5x-2 as \left(12x^{2}-3x\right)+\left(8x-2\right).

$$\left(12x^{2}-3x\right)+\left(8x-2\right)$$

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

$$3x\left(4x-1\right)+2\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(3x+2\right)$$

Rewrite the complete factored expression.

$$3\left(4x-1\right)\left(3x+2\right)$$

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.

$$36x^{2}+15x-6=0$$

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.

$$x=\frac{-15±\sqrt{15^{2}-4\times 36\left(-6\right)}}{2\times 36}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\times 36\left(-6\right)}}{2\times 36}$$

Multiply -4 times 36.

$$x=\frac{-15±\sqrt{225-144\left(-6\right)}}{2\times 36}$$

Multiply -144 times -6.

$$x=\frac{-15±\sqrt{225+864}}{2\times 36}$$

Add 225 to 864.

$$x=\frac{-15±\sqrt{1089}}{2\times 36}$$

Take the square root of 1089.

$$x=\frac{-15±33}{2\times 36}$$

Multiply 2 times 36.

$$x=\frac{-15±33}{72}$$

Now solve the equation x=\frac{-15±33}{72} when ± is plus. Add -15 to 33.

$$x=\frac{18}{72}$$

Reduce the fraction \frac{18}{72} to lowest terms by extracting and canceling out 18.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-15±33}{72} when ± is minus. Subtract 33 from -15.

$$x=-\frac{48}{72}$$

Reduce the fraction \frac{-48}{72} to lowest terms by extracting and canceling out 24.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{1}{4} for x_{1} and -\frac{2}{3} for x_{2}.

$$36x^{2}+15x-6=36\left(x-\frac{1}{4}\right)\left(x-\left(-\frac{2}{3}\right)\right)$$

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

$$36x^{2}+15x-6=36\left(x-\frac{1}{4}\right)\left(x+\frac{2}{3}\right)$$

Subtract \frac{1}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$36x^{2}+15x-6=36\times \frac{4x-1}{4}\left(x+\frac{2}{3}\right)$$

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

$$36x^{2}+15x-6=36\times \frac{4x-1}{4}\times \frac{3x+2}{3}$$

Multiply \frac{4x-1}{4} times \frac{3x+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$36x^{2}+15x-6=36\times \frac{\left(4x-1\right)\left(3x+2\right)}{4\times 3}$$

Multiply 4 times 3.

$$36x^{2}+15x-6=36\times \frac{\left(4x-1\right)\left(3x+2\right)}{12}$$

Cancel out 12, the greatest common factor in 36 and 12.

$$36x^{2}+15x-6=3\left(4x-1\right)\left(3x+2\right)$$

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 36

$$x ^ 2 +\frac{5}{12}x -\frac{1}{6} = 0$$

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

$$r + s = -\frac{5}{12} rs = -\frac{1}{6}$$

Two numbers r and s sum up to -\frac{5}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{12} = -\frac{5}{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>

$$r = -\frac{5}{24} - u s = -\frac{5}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{6}

$$(-\frac{5}{24} - u) (-\frac{5}{24} + u) = -\frac{1}{6}$$

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

$$\frac{25}{576} - u^2 = -\frac{1}{6}$$

Simplify the expression by subtracting \frac{25}{576} on both sides

$$-u^2 = -\frac{1}{6}-\frac{25}{576} = -\frac{121}{576}$$

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

$$u^2 = \frac{121}{576} u = \pm\sqrt{\frac{121}{576}} = \pm \frac{11}{24}$$

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

$$r =-\frac{5}{24} - \frac{11}{24} = -0.667 s = -\frac{5}{24} + \frac{11}{24} = 0.250$$

Evaluate

$$36x^{2}+15x-6$$