Solve 3c^2+1c-24

$$3 c ^ { 2 } + 1 c - 24$$

Factor

$$\left(3c-8\right)\left(c+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3c^{2}+ac+bc-24. To find a and b, set up a system to be solved.

$$a+b=1 ab=3\left(-24\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$a=-8 b=9$$

Rewrite 3c^{2}+c-24 as \left(3c^{2}-8c\right)+\left(9c-24\right).

$$\left(3c^{2}-8c\right)+\left(9c-24\right)$$

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

$$c\left(3c-8\right)+3\left(3c-8\right)$$

Factor out common term 3c-8 by using distributive property.

$$\left(3c-8\right)\left(c+3\right)$$

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.

$$3c^{2}+c-24=0$$

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.

$$c=\frac{-1±\sqrt{1^{2}-4\times 3\left(-24\right)}}{2\times 3}$$

Square 1.

$$c=\frac{-1±\sqrt{1-4\times 3\left(-24\right)}}{2\times 3}$$

Multiply -4 times 3.

$$c=\frac{-1±\sqrt{1-12\left(-24\right)}}{2\times 3}$$

Multiply -12 times -24.

$$c=\frac{-1±\sqrt{1+288}}{2\times 3}$$

Add 1 to 288.

$$c=\frac{-1±\sqrt{289}}{2\times 3}$$

Take the square root of 289.

$$c=\frac{-1±17}{2\times 3}$$

Multiply 2 times 3.

$$c=\frac{-1±17}{6}$$

Now solve the equation c=\frac{-1±17}{6} when ± is plus. Add -1 to 17.

$$c=\frac{16}{6}$$

Reduce the fraction \frac{16}{6} to lowest terms by extracting and canceling out 2.

$$c=\frac{8}{3}$$

Now solve the equation c=\frac{-1±17}{6} when ± is minus. Subtract 17 from -1.

$$c=-\frac{18}{6}$$

Divide -18 by 6.

$$c=-3$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{8}{3} for x_{1} and -3 for x_{2}.

$$3c^{2}+c-24=3\left(c-\frac{8}{3}\right)\left(c-\left(-3\right)\right)$$

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

$$3c^{2}+c-24=3\left(c-\frac{8}{3}\right)\left(c+3\right)$$

Subtract \frac{8}{3} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$3c^{2}+c-24=3\times \frac{3c-8}{3}\left(c+3\right)$$

Cancel out 3, the greatest common factor in 3 and 3.

$$3c^{2}+c-24=\left(3c-8\right)\left(c+3\right)$$

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 3

$$x ^ 2 +\frac{1}{3}x -8 = 0$$

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

$$r + s = -\frac{1}{3} rs = -8$$

Two numbers r and s sum up to -\frac{1}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{1}{3} = -\frac{1}{6}. 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>

$$r = -\frac{1}{6} - u s = -\frac{1}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -8

$$(-\frac{1}{6} - u) (-\frac{1}{6} + u) = -8$$

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

$$\frac{1}{36} - u^2 = -8$$

Simplify the expression by subtracting \frac{1}{36} on both sides

$$-u^2 = -8-\frac{1}{36} = -\frac{289}{36}$$

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

$$u^2 = \frac{289}{36} u = \pm\sqrt{\frac{289}{36}} = \pm \frac{17}{6}$$

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

$$r =-\frac{1}{6} - \frac{17}{6} = -3 s = -\frac{1}{6} + \frac{17}{6} = 2.667$$

Evaluate

$$\left(3c-8\right)\left(c+3\right)$$