Solve 3c^2-2c-5

$$3 c ^ { 2 } - 2 c - 5$$

Factor

$$\left(3c-5\right)\left(c+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 3c^{2}+ac+bc-5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-5\right)=-15$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -15.

$$1,-15 3,-5$$

Calculate the sum for each pair.

$$1-15=-14 3-5=-2$$

The solution is the pair that gives sum -2.

$$a=-5 b=3$$

Rewrite 3c^{2}-2c-5 as \left(3c^{2}-5c\right)+\left(3c-5\right).

$$\left(3c^{2}-5c\right)+\left(3c-5\right)$$

Factor out c in 3c^{2}-5c.

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

Factor out common term 3c-5 by using distributive property.

$$\left(3c-5\right)\left(c+1\right)$$

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.

$$3c^{2}-2c-5=0$$

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.

$$c=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-5\right)}}{2\times 3}$$

Square -2.

$$c=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-5\right)}}{2\times 3}$$

Multiply -4 times 3.

$$c=\frac{-\left(-2\right)±\sqrt{4-12\left(-5\right)}}{2\times 3}$$

Multiply -12 times -5.

$$c=\frac{-\left(-2\right)±\sqrt{4+60}}{2\times 3}$$

Add 4 to 60.

$$c=\frac{-\left(-2\right)±\sqrt{64}}{2\times 3}$$

Take the square root of 64.

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

The opposite of -2 is 2.

$$c=\frac{2±8}{2\times 3}$$

Multiply 2 times 3.

$$c=\frac{2±8}{6}$$

Now solve the equation c=\frac{2±8}{6} when ± is plus. Add 2 to 8.

$$c=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation c=\frac{2±8}{6} when ± is minus. Subtract 8 from 2.

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

Divide -6 by 6.

$$c=-1$$

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

$$3c^{2}-2c-5=3\left(c-\frac{5}{3}\right)\left(c-\left(-1\right)\right)$$

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

$$3c^{2}-2c-5=3\left(c-\frac{5}{3}\right)\left(c+1\right)$$

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

$$3c^{2}-2c-5=3\times \frac{3c-5}{3}\left(c+1\right)$$

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

$$3c^{2}-2c-5=\left(3c-5\right)\left(c+1\right)$$

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

$$x ^ 2 -\frac{2}{3}x -\frac{5}{3} = 0$$

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

$$r + s = \frac{2}{3} rs = -\frac{5}{3}$$

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

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{3}

$$(\frac{1}{3} - u) (\frac{1}{3} + u) = -\frac{5}{3}$$

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

$$\frac{1}{9} - u^2 = -\frac{5}{3}$$

Simplify the expression by subtracting \frac{1}{9} on both sides

$$-u^2 = -\frac{5}{3}-\frac{1}{9} = -\frac{16}{9}$$

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

$$u^2 = \frac{16}{9} u = \pm\sqrt{\frac{16}{9}} = \pm \frac{4}{3}$$

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

$$r =\frac{1}{3} - \frac{4}{3} = -1.000 s = \frac{1}{3} + \frac{4}{3} = 1.667$$

Evaluate

$$\left(3c-5\right)\left(c+1\right)$$