Solve 12k^2+15k-27

$$12 k ^ { 2 } + 15 k - 27$$

Factor

$$3\left(k-1\right)\left(4k+9\right)$$

Solution Steps

Factor out 3.

$$3\left(4k^{2}+5k-9\right)$$

Consider 4k^{2}+5k-9. Factor the expression by grouping. First, the expression needs to be rewritten as 4k^{2}+ak+bk-9. To find a and b, set up a system to be solved.

$$a+b=5 ab=4\left(-9\right)=-36$$

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 5.

$$a=-4 b=9$$

Rewrite 4k^{2}+5k-9 as \left(4k^{2}-4k\right)+\left(9k-9\right).

$$\left(4k^{2}-4k\right)+\left(9k-9\right)$$

Factor out 4k in the first and 9 in the second group.

$$4k\left(k-1\right)+9\left(k-1\right)$$

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

$$\left(k-1\right)\left(4k+9\right)$$

Rewrite the complete factored expression.

$$3\left(k-1\right)\left(4k+9\right)$$

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.

$$12k^{2}+15k-27=0$$

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.

$$k=\frac{-15±\sqrt{15^{2}-4\times 12\left(-27\right)}}{2\times 12}$$

Square 15.

$$k=\frac{-15±\sqrt{225-4\times 12\left(-27\right)}}{2\times 12}$$

Multiply -4 times 12.

$$k=\frac{-15±\sqrt{225-48\left(-27\right)}}{2\times 12}$$

Multiply -48 times -27.

$$k=\frac{-15±\sqrt{225+1296}}{2\times 12}$$

Add 225 to 1296.

$$k=\frac{-15±\sqrt{1521}}{2\times 12}$$

Take the square root of 1521.

$$k=\frac{-15±39}{2\times 12}$$

Multiply 2 times 12.

$$k=\frac{-15±39}{24}$$

Now solve the equation k=\frac{-15±39}{24} when ± is plus. Add -15 to 39.

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

Divide 24 by 24.

$$k=1$$

Now solve the equation k=\frac{-15±39}{24} when ± is minus. Subtract 39 from -15.

$$k=-\frac{54}{24}$$

Reduce the fraction \frac{-54}{24} to lowest terms by extracting and canceling out 6.

$$k=-\frac{9}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 1 for x_{1} and -\frac{9}{4} for x_{2}.

$$12k^{2}+15k-27=12\left(k-1\right)\left(k-\left(-\frac{9}{4}\right)\right)$$

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

$$12k^{2}+15k-27=12\left(k-1\right)\left(k+\frac{9}{4}\right)$$

Add \frac{9}{4} to k by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$12k^{2}+15k-27=12\left(k-1\right)\times \frac{4k+9}{4}$$

Cancel out 4, the greatest common factor in 12 and 4.

$$12k^{2}+15k-27=3\left(k-1\right)\left(4k+9\right)$$

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

$$x ^ 2 +\frac{5}{4}x -\frac{9}{4} = 0$$

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

$$r + s = -\frac{5}{4} rs = -\frac{9}{4}$$

Two numbers r and s sum up to -\frac{5}{4} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{4} = -\frac{5}{8}. 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>

$$r = -\frac{5}{8} - u s = -\frac{5}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{4}

$$(-\frac{5}{8} - u) (-\frac{5}{8} + u) = -\frac{9}{4}$$

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

$$\frac{25}{64} - u^2 = -\frac{9}{4}$$

Simplify the expression by subtracting \frac{25}{64} on both sides

$$-u^2 = -\frac{9}{4}-\frac{25}{64} = -\frac{169}{64}$$

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

$$u^2 = \frac{169}{64} u = \pm\sqrt{\frac{169}{64}} = \pm \frac{13}{8}$$

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

$$r =-\frac{5}{8} - \frac{13}{8} = -2.250 s = -\frac{5}{8} + \frac{13}{8} = 1$$

Evaluate

$$3\left(k-1\right)\left(4k+9\right)$$