Solve 12c^2+11c-15

$$12 c ^ { 2 } + 11 c - 15$$

Factor

$$\left(4c-3\right)\left(3c+5\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 12c^{2}+ac+bc-15. To find a and b, set up a system to be solved.

$$a+b=11 ab=12\left(-15\right)=-180$$

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 -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 11.

$$a=-9 b=20$$

Rewrite 12c^{2}+11c-15 as \left(12c^{2}-9c\right)+\left(20c-15\right).

$$\left(12c^{2}-9c\right)+\left(20c-15\right)$$

Factor out 3c in the first and 5 in the second group.

$$3c\left(4c-3\right)+5\left(4c-3\right)$$

Factor out common term 4c-3 by using distributive property.

$$\left(4c-3\right)\left(3c+5\right)$$

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.

$$12c^{2}+11c-15=0$$

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.

$$c=\frac{-11±\sqrt{11^{2}-4\times 12\left(-15\right)}}{2\times 12}$$

Square 11.

$$c=\frac{-11±\sqrt{121-4\times 12\left(-15\right)}}{2\times 12}$$

Multiply -4 times 12.

$$c=\frac{-11±\sqrt{121-48\left(-15\right)}}{2\times 12}$$

Multiply -48 times -15.

$$c=\frac{-11±\sqrt{121+720}}{2\times 12}$$

Add 121 to 720.

$$c=\frac{-11±\sqrt{841}}{2\times 12}$$

Take the square root of 841.

$$c=\frac{-11±29}{2\times 12}$$

Multiply 2 times 12.

$$c=\frac{-11±29}{24}$$

Now solve the equation c=\frac{-11±29}{24} when ± is plus. Add -11 to 29.

$$c=\frac{18}{24}$$

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

$$c=\frac{3}{4}$$

Now solve the equation c=\frac{-11±29}{24} when ± is minus. Subtract 29 from -11.

$$c=-\frac{40}{24}$$

Reduce the fraction \frac{-40}{24} to lowest terms by extracting and canceling out 8.

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

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{5}{3} for x_{2}.

$$12c^{2}+11c-15=12\left(c-\frac{3}{4}\right)\left(c-\left(-\frac{5}{3}\right)\right)$$

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

$$12c^{2}+11c-15=12\left(c-\frac{3}{4}\right)\left(c+\frac{5}{3}\right)$$

Subtract \frac{3}{4} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$12c^{2}+11c-15=12\times \frac{4c-3}{4}\left(c+\frac{5}{3}\right)$$

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

$$12c^{2}+11c-15=12\times \frac{4c-3}{4}\times \frac{3c+5}{3}$$

Multiply \frac{4c-3}{4} times \frac{3c+5}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$12c^{2}+11c-15=12\times \frac{\left(4c-3\right)\left(3c+5\right)}{4\times 3}$$

Multiply 4 times 3.

$$12c^{2}+11c-15=12\times \frac{\left(4c-3\right)\left(3c+5\right)}{12}$$

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

$$12c^{2}+11c-15=\left(4c-3\right)\left(3c+5\right)$$

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

$$x ^ 2 +\frac{11}{12}x -\frac{5}{4} = 0$$

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

$$r + s = -\frac{11}{12} rs = -\frac{5}{4}$$

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 = -\frac{5}{4}

$$(-\frac{11}{24} - u) (-\frac{11}{24} + u) = -\frac{5}{4}$$

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

$$\frac{121}{576} - u^2 = -\frac{5}{4}$$

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

$$-u^2 = -\frac{5}{4}-\frac{121}{576} = -\frac{841}{576}$$

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

$$u^2 = \frac{841}{576} u = \pm\sqrt{\frac{841}{576}} = \pm \frac{29}{24}$$

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

$$r =-\frac{11}{24} - \frac{29}{24} = -1.667 s = -\frac{11}{24} + \frac{29}{24} = 0.750$$

Evaluate

$$\left(4c-3\right)\left(3c+5\right)$$