Solve 12k^2+25k+12

$$12 k ^ { 2 } + 25 k + 12$$

Factor

$$\left(3k+4\right)\left(4k+3\right)$$

Solution Steps

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

$$a+b=25 ab=12\times 12=144$$

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 144.

$$1,144 2,72 3,48 4,36 6,24 8,18 9,16 12,12$$

Calculate the sum for each pair.

$$1+144=145 2+72=74 3+48=51 4+36=40 6+24=30 8+18=26 9+16=25 12+12=24$$

The solution is the pair that gives sum 25.

$$a=9 b=16$$

Rewrite 12k^{2}+25k+12 as \left(12k^{2}+9k\right)+\left(16k+12\right).

$$\left(12k^{2}+9k\right)+\left(16k+12\right)$$

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

$$3k\left(4k+3\right)+4\left(4k+3\right)$$

Factor out common term 4k+3 by using distributive property.

$$\left(4k+3\right)\left(3k+4\right)$$

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.

$$12k^{2}+25k+12=0$$

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.

$$k=\frac{-25±\sqrt{25^{2}-4\times 12\times 12}}{2\times 12}$$

Square 25.

$$k=\frac{-25±\sqrt{625-4\times 12\times 12}}{2\times 12}$$

Multiply -4 times 12.

$$k=\frac{-25±\sqrt{625-48\times 12}}{2\times 12}$$

Multiply -48 times 12.

$$k=\frac{-25±\sqrt{625-576}}{2\times 12}$$

Add 625 to -576.

$$k=\frac{-25±\sqrt{49}}{2\times 12}$$

Take the square root of 49.

$$k=\frac{-25±7}{2\times 12}$$

Multiply 2 times 12.

$$k=\frac{-25±7}{24}$$

Now solve the equation k=\frac{-25±7}{24} when ± is plus. Add -25 to 7.

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

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

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

Now solve the equation k=\frac{-25±7}{24} when ± is minus. Subtract 7 from -25.

$$k=-\frac{32}{24}$$

Reduce the fraction \frac{-32}{24} to lowest terms by extracting and canceling out 8.

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

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

$$12k^{2}+25k+12=12\left(k-\left(-\frac{3}{4}\right)\right)\left(k-\left(-\frac{4}{3}\right)\right)$$

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

$$12k^{2}+25k+12=12\left(k+\frac{3}{4}\right)\left(k+\frac{4}{3}\right)$$

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

$$12k^{2}+25k+12=12\times \frac{4k+3}{4}\left(k+\frac{4}{3}\right)$$

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

$$12k^{2}+25k+12=12\times \frac{4k+3}{4}\times \frac{3k+4}{3}$$

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

$$12k^{2}+25k+12=12\times \frac{\left(4k+3\right)\left(3k+4\right)}{4\times 3}$$

Multiply 4 times 3.

$$12k^{2}+25k+12=12\times \frac{\left(4k+3\right)\left(3k+4\right)}{12}$$

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

$$12k^{2}+25k+12=\left(4k+3\right)\left(3k+4\right)$$

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 12

$$x ^ 2 +\frac{25}{12}x +1 = 0$$

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

$$r + s = -\frac{25}{12} rs = 1$$

Two numbers r and s sum up to -\frac{25}{12} exactly when the average of the two numbers is \frac{1}{2}*-\frac{25}{12} = -\frac{25}{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>

$$r = -\frac{25}{24} - u s = -\frac{25}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = 1

$$(-\frac{25}{24} - u) (-\frac{25}{24} + u) = 1$$

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

$$\frac{625}{576} - u^2 = 1$$

Simplify the expression by subtracting \frac{625}{576} on both sides

$$-u^2 = 1-\frac{625}{576} = -\frac{49}{576}$$

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

$$u^2 = \frac{49}{576} u = \pm\sqrt{\frac{49}{576}} = \pm \frac{7}{24}$$

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

$$r =-\frac{25}{24} - \frac{7}{24} = -1.333 s = -\frac{25}{24} + \frac{7}{24} = -0.750$$

Evaluate

$$\left(3k+4\right)\left(4k+3\right)$$