Solve 30k^2+38k+12

$$30 k ^ { 2 } + 38 k + 12$$

Factor

$$2\left(3k+2\right)\left(5k+3\right)$$

Solution Steps

Factor out 2.

$$2\left(15k^{2}+19k+6\right)$$

Consider 15k^{2}+19k+6. Factor the expression by grouping. First, the expression needs to be rewritten as 15k^{2}+ak+bk+6. To find a and b, set up a system to be solved.

$$a+b=19 ab=15\times 6=90$$

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 90.

$$1,90 2,45 3,30 5,18 6,15 9,10$$

Calculate the sum for each pair.

$$1+90=91 2+45=47 3+30=33 5+18=23 6+15=21 9+10=19$$

The solution is the pair that gives sum 19.

$$a=9 b=10$$

Rewrite 15k^{2}+19k+6 as \left(15k^{2}+9k\right)+\left(10k+6\right).

$$\left(15k^{2}+9k\right)+\left(10k+6\right)$$

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

$$3k\left(5k+3\right)+2\left(5k+3\right)$$

Factor out common term 5k+3 by using distributive property.

$$\left(5k+3\right)\left(3k+2\right)$$

Rewrite the complete factored expression.

$$2\left(5k+3\right)\left(3k+2\right)$$

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.

$$30k^{2}+38k+12=0$$

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.

$$k=\frac{-38±\sqrt{38^{2}-4\times 30\times 12}}{2\times 30}$$

Square 38.

$$k=\frac{-38±\sqrt{1444-4\times 30\times 12}}{2\times 30}$$

Multiply -4 times 30.

$$k=\frac{-38±\sqrt{1444-120\times 12}}{2\times 30}$$

Multiply -120 times 12.

$$k=\frac{-38±\sqrt{1444-1440}}{2\times 30}$$

Add 1444 to -1440.

$$k=\frac{-38±\sqrt{4}}{2\times 30}$$

Take the square root of 4.

$$k=\frac{-38±2}{2\times 30}$$

Multiply 2 times 30.

$$k=\frac{-38±2}{60}$$

Now solve the equation k=\frac{-38±2}{60} when ± is plus. Add -38 to 2.

$$k=-\frac{36}{60}$$

Reduce the fraction \frac{-36}{60} to lowest terms by extracting and canceling out 12.

$$k=-\frac{3}{5}$$

Now solve the equation k=\frac{-38±2}{60} when ± is minus. Subtract 2 from -38.

$$k=-\frac{40}{60}$$

Reduce the fraction \frac{-40}{60} to lowest terms by extracting and canceling out 20.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{5} for x_{1} and -\frac{2}{3} for x_{2}.

$$30k^{2}+38k+12=30\left(k-\left(-\frac{3}{5}\right)\right)\left(k-\left(-\frac{2}{3}\right)\right)$$

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

$$30k^{2}+38k+12=30\left(k+\frac{3}{5}\right)\left(k+\frac{2}{3}\right)$$

Add \frac{3}{5} to k by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$30k^{2}+38k+12=30\times \frac{5k+3}{5}\left(k+\frac{2}{3}\right)$$

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

$$30k^{2}+38k+12=30\times \frac{5k+3}{5}\times \frac{3k+2}{3}$$

Multiply \frac{5k+3}{5} times \frac{3k+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$30k^{2}+38k+12=30\times \frac{\left(5k+3\right)\left(3k+2\right)}{5\times 3}$$

Multiply 5 times 3.

$$30k^{2}+38k+12=30\times \frac{\left(5k+3\right)\left(3k+2\right)}{15}$$

Cancel out 15, the greatest common factor in 30 and 15.

$$30k^{2}+38k+12=2\left(5k+3\right)\left(3k+2\right)$$

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 30

$$x ^ 2 +\frac{19}{15}x +\frac{2}{5} = 0$$

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

$$r + s = -\frac{19}{15} rs = \frac{2}{5}$$

Two numbers r and s sum up to -\frac{19}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{15} = -\frac{19}{30}. 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>

$$r = -\frac{19}{30} - u s = -\frac{19}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{5}

$$(-\frac{19}{30} - u) (-\frac{19}{30} + u) = \frac{2}{5}$$

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

$$\frac{361}{900} - u^2 = \frac{2}{5}$$

Simplify the expression by subtracting \frac{361}{900} on both sides

$$-u^2 = \frac{2}{5}-\frac{361}{900} = -\frac{1}{900}$$

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

$$u^2 = \frac{1}{900} u = \pm\sqrt{\frac{1}{900}} = \pm \frac{1}{30}$$

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

$$r =-\frac{19}{30} - \frac{1}{30} = -0.667 s = -\frac{19}{30} + \frac{1}{30} = -0.600$$

Evaluate

$$30k^{2}+38k+12$$