Solve 12a^2-11a-36

$$12 a ^ { 2 } - 11 a - 36$$

Factor

$$\left(4a-9\right)\left(3a+4\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12a^{2}+pa+qa-36. To find p and q, set up a system to be solved.

$$p+q=-11 pq=12\left(-36\right)=-432$$

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -432.

$$1,-432 2,-216 3,-144 4,-108 6,-72 8,-54 9,-48 12,-36 16,-27 18,-24$$

Calculate the sum for each pair.

$$1-432=-431 2-216=-214 3-144=-141 4-108=-104 6-72=-66 8-54=-46 9-48=-39 12-36=-24 16-27=-11 18-24=-6$$

The solution is the pair that gives sum -11.

$$p=-27 q=16$$

Rewrite 12a^{2}-11a-36 as \left(12a^{2}-27a\right)+\left(16a-36\right).

$$\left(12a^{2}-27a\right)+\left(16a-36\right)$$

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

$$3a\left(4a-9\right)+4\left(4a-9\right)$$

Factor out common term 4a-9 by using distributive property.

$$\left(4a-9\right)\left(3a+4\right)$$

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.

$$12a^{2}-11a-36=0$$

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.

$$a=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 12\left(-36\right)}}{2\times 12}$$

Square -11.

$$a=\frac{-\left(-11\right)±\sqrt{121-4\times 12\left(-36\right)}}{2\times 12}$$

Multiply -4 times 12.

$$a=\frac{-\left(-11\right)±\sqrt{121-48\left(-36\right)}}{2\times 12}$$

Multiply -48 times -36.

$$a=\frac{-\left(-11\right)±\sqrt{121+1728}}{2\times 12}$$

Add 121 to 1728.

$$a=\frac{-\left(-11\right)±\sqrt{1849}}{2\times 12}$$

Take the square root of 1849.

$$a=\frac{-\left(-11\right)±43}{2\times 12}$$

The opposite of -11 is 11.

$$a=\frac{11±43}{2\times 12}$$

Multiply 2 times 12.

$$a=\frac{11±43}{24}$$

Now solve the equation a=\frac{11±43}{24} when ± is plus. Add 11 to 43.

$$a=\frac{54}{24}$$

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

$$a=\frac{9}{4}$$

Now solve the equation a=\frac{11±43}{24} when ± is minus. Subtract 43 from 11.

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

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

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{9}{4} for x_{1} and -\frac{4}{3} for x_{2}.

$$12a^{2}-11a-36=12\left(a-\frac{9}{4}\right)\left(a-\left(-\frac{4}{3}\right)\right)$$

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

$$12a^{2}-11a-36=12\left(a-\frac{9}{4}\right)\left(a+\frac{4}{3}\right)$$

Subtract \frac{9}{4} from a by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-11a-36=12\times \frac{4a-9}{4}\left(a+\frac{4}{3}\right)$$

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

$$12a^{2}-11a-36=12\times \frac{4a-9}{4}\times \frac{3a+4}{3}$$

Multiply \frac{4a-9}{4} times \frac{3a+4}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12a^{2}-11a-36=12\times \frac{\left(4a-9\right)\left(3a+4\right)}{4\times 3}$$

Multiply 4 times 3.

$$12a^{2}-11a-36=12\times \frac{\left(4a-9\right)\left(3a+4\right)}{12}$$

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

$$12a^{2}-11a-36=\left(4a-9\right)\left(3a+4\right)$$

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

$$x ^ 2 -\frac{11}{12}x -3 = 0$$

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

$$r + s = \frac{11}{12} rs = -3$$

Two numbers r and s sum up to \frac{11}{12} exactly when the average of the two numbers is \frac{1}{2}*\frac{11}{12} = \frac{11}{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>

$$r = \frac{11}{24} - u s = \frac{11}{24} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -3

$$(\frac{11}{24} - u) (\frac{11}{24} + u) = -3$$

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

$$\frac{121}{576} - u^2 = -3$$

Simplify the expression by subtracting \frac{121}{576} on both sides

$$-u^2 = -3-\frac{121}{576} = -\frac{1849}{576}$$

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

$$u^2 = \frac{1849}{576} u = \pm\sqrt{\frac{1849}{576}} = \pm \frac{43}{24}$$

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

$$r =\frac{11}{24} - \frac{43}{24} = -1.333 s = \frac{11}{24} + \frac{43}{24} = 2.250$$

Evaluate

$$\left(4a-9\right)\left(3a+4\right)$$