Solve 12k^2+16k-3

$$12 k ^ { 2 } + 16 k - 3$$

Factor

$$\left(6k-1\right)\left(2k+3\right)$$

Solution Steps

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

$$a+b=16 ab=12\left(-3\right)=-36$$

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 -36.

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

Calculate the sum for each pair.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

The solution is the pair that gives sum 16.

$$a=-2 b=18$$

Rewrite 12k^{2}+16k-3 as \left(12k^{2}-2k\right)+\left(18k-3\right).

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

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

$$2k\left(6k-1\right)+3\left(6k-1\right)$$

Factor out common term 6k-1 by using distributive property.

$$\left(6k-1\right)\left(2k+3\right)$$

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.

$$12k^{2}+16k-3=0$$

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.

$$k=\frac{-16±\sqrt{16^{2}-4\times 12\left(-3\right)}}{2\times 12}$$

Square 16.

$$k=\frac{-16±\sqrt{256-4\times 12\left(-3\right)}}{2\times 12}$$

Multiply -4 times 12.

$$k=\frac{-16±\sqrt{256-48\left(-3\right)}}{2\times 12}$$

Multiply -48 times -3.

$$k=\frac{-16±\sqrt{256+144}}{2\times 12}$$

Add 256 to 144.

$$k=\frac{-16±\sqrt{400}}{2\times 12}$$

Take the square root of 400.

$$k=\frac{-16±20}{2\times 12}$$

Multiply 2 times 12.

$$k=\frac{-16±20}{24}$$

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

$$k=\frac{4}{24}$$

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

$$k=\frac{1}{6}$$

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

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

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

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

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}.

$$12k^{2}+16k-3=12\left(k-\frac{1}{6}\right)\left(k-\left(-\frac{3}{2}\right)\right)$$

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

$$12k^{2}+16k-3=12\left(k-\frac{1}{6}\right)\left(k+\frac{3}{2}\right)$$

Subtract \frac{1}{6} from k by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12k^{2}+16k-3=12\times \frac{6k-1}{6}\left(k+\frac{3}{2}\right)$$

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

$$12k^{2}+16k-3=12\times \frac{6k-1}{6}\times \frac{2k+3}{2}$$

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

$$12k^{2}+16k-3=12\times \frac{\left(6k-1\right)\left(2k+3\right)}{6\times 2}$$

Multiply 6 times 2.

$$12k^{2}+16k-3=12\times \frac{\left(6k-1\right)\left(2k+3\right)}{12}$$

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

$$12k^{2}+16k-3=\left(6k-1\right)\left(2k+3\right)$$

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

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

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

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

Two numbers r and s sum up to -\frac{4}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{4}{3} = -\frac{2}{3}. 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>

$$r = -\frac{2}{3} - u s = -\frac{2}{3} + u$$

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

$$(-\frac{2}{3} - u) (-\frac{2}{3} + u) = -\frac{1}{4}$$

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

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

Simplify the expression by subtracting \frac{4}{9} on both sides

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

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

$$u^2 = \frac{25}{36} u = \pm\sqrt{\frac{25}{36}} = \pm \frac{5}{6}$$

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

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

Evaluate

$$\left(6k-1\right)\left(2k+3\right)$$