Solve 10c^2-19c-15

$$10 c ^ { 2 } - 19 c - 15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10c^{2}+ac+bc-15. To find a and b, set up a system to be solved.

$$a+b=-19 ab=10\left(-15\right)=-150$$

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 -150.

$$1,-150 2,-75 3,-50 5,-30 6,-25 10,-15$$

Calculate the sum for each pair.

$$1-150=-149 2-75=-73 3-50=-47 5-30=-25 6-25=-19 10-15=-5$$

The solution is the pair that gives sum -19.

$$a=-25 b=6$$

Rewrite 10c^{2}-19c-15 as \left(10c^{2}-25c\right)+\left(6c-15\right).

$$\left(10c^{2}-25c\right)+\left(6c-15\right)$$

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

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

Factor out common term 2c-5 by using distributive property.

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

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.

$$10c^{2}-19c-15=0$$

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.

$$c=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 10\left(-15\right)}}{2\times 10}$$

Square -19.

$$c=\frac{-\left(-19\right)±\sqrt{361-4\times 10\left(-15\right)}}{2\times 10}$$

Multiply -4 times 10.

$$c=\frac{-\left(-19\right)±\sqrt{361-40\left(-15\right)}}{2\times 10}$$

Multiply -40 times -15.

$$c=\frac{-\left(-19\right)±\sqrt{361+600}}{2\times 10}$$

Add 361 to 600.

$$c=\frac{-\left(-19\right)±\sqrt{961}}{2\times 10}$$

Take the square root of 961.

$$c=\frac{-\left(-19\right)±31}{2\times 10}$$

The opposite of -19 is 19.

$$c=\frac{19±31}{2\times 10}$$

Multiply 2 times 10.

$$c=\frac{19±31}{20}$$

Now solve the equation c=\frac{19±31}{20} when ± is plus. Add 19 to 31.

$$c=\frac{50}{20}$$

Reduce the fraction \frac{50}{20} to lowest terms by extracting and canceling out 10.

$$c=\frac{5}{2}$$

Now solve the equation c=\frac{19±31}{20} when ± is minus. Subtract 31 from 19.

$$c=-\frac{12}{20}$$

Reduce the fraction \frac{-12}{20} to lowest terms by extracting and canceling out 4.

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

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

$$10c^{2}-19c-15=10\left(c-\frac{5}{2}\right)\left(c-\left(-\frac{3}{5}\right)\right)$$

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

$$10c^{2}-19c-15=10\left(c-\frac{5}{2}\right)\left(c+\frac{3}{5}\right)$$

Subtract \frac{5}{2} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10c^{2}-19c-15=10\times \frac{2c-5}{2}\left(c+\frac{3}{5}\right)$$

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

$$10c^{2}-19c-15=10\times \frac{2c-5}{2}\times \frac{5c+3}{5}$$

Multiply \frac{2c-5}{2} times \frac{5c+3}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10c^{2}-19c-15=10\times \frac{\left(2c-5\right)\left(5c+3\right)}{2\times 5}$$

Multiply 2 times 5.

$$10c^{2}-19c-15=10\times \frac{\left(2c-5\right)\left(5c+3\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10c^{2}-19c-15=\left(2c-5\right)\left(5c+3\right)$$

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 10

$$x ^ 2 -\frac{19}{10}x -\frac{3}{2} = 0$$

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

$$r + s = \frac{19}{10} rs = -\frac{3}{2}$$

Two numbers r and s sum up to \frac{19}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{19}{10} = \frac{19}{20}. 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>

$$r = \frac{19}{20} - u s = \frac{19}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{2}

$$(\frac{19}{20} - u) (\frac{19}{20} + u) = -\frac{3}{2}$$

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

$$\frac{361}{400} - u^2 = -\frac{3}{2}$$

Simplify the expression by subtracting \frac{361}{400} on both sides

$$-u^2 = -\frac{3}{2}-\frac{361}{400} = -\frac{961}{400}$$

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

$$u^2 = \frac{961}{400} u = \pm\sqrt{\frac{961}{400}} = \pm \frac{31}{20}$$

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

$$r =\frac{19}{20} - \frac{31}{20} = -0.600 s = \frac{19}{20} + \frac{31}{20} = 2.500$$

Evaluate

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