Solve 20c^2+13c-15

$$20 c ^ { 2 } + 13 c - 15$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 20c^{2}+ac+bc-15. To find a and b, set up a system to be solved.

$$a+b=13 ab=20\left(-15\right)=-300$$

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 -300.

$$-1,300 -2,150 -3,100 -4,75 -5,60 -6,50 -10,30 -12,25 -15,20$$

Calculate the sum for each pair.

$$-1+300=299 -2+150=148 -3+100=97 -4+75=71 -5+60=55 -6+50=44 -10+30=20 -12+25=13 -15+20=5$$

The solution is the pair that gives sum 13.

$$a=-12 b=25$$

Rewrite 20c^{2}+13c-15 as \left(20c^{2}-12c\right)+\left(25c-15\right).

$$\left(20c^{2}-12c\right)+\left(25c-15\right)$$

Factor out 4c in the first and 5 in the second group.

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

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

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

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.

$$20c^{2}+13c-15=0$$

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.

$$c=\frac{-13±\sqrt{13^{2}-4\times 20\left(-15\right)}}{2\times 20}$$

Square 13.

$$c=\frac{-13±\sqrt{169-4\times 20\left(-15\right)}}{2\times 20}$$

Multiply -4 times 20.

$$c=\frac{-13±\sqrt{169-80\left(-15\right)}}{2\times 20}$$

Multiply -80 times -15.

$$c=\frac{-13±\sqrt{169+1200}}{2\times 20}$$

Add 169 to 1200.

$$c=\frac{-13±\sqrt{1369}}{2\times 20}$$

Take the square root of 1369.

$$c=\frac{-13±37}{2\times 20}$$

Multiply 2 times 20.

$$c=\frac{-13±37}{40}$$

Now solve the equation c=\frac{-13±37}{40} when ± is plus. Add -13 to 37.

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

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

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

Now solve the equation c=\frac{-13±37}{40} when ± is minus. Subtract 37 from -13.

$$c=-\frac{50}{40}$$

Reduce the fraction \frac{-50}{40} to lowest terms by extracting and canceling out 10.

$$c=-\frac{5}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{5} for x_{1} and -\frac{5}{4} for x_{2}.

$$20c^{2}+13c-15=20\left(c-\frac{3}{5}\right)\left(c-\left(-\frac{5}{4}\right)\right)$$

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

$$20c^{2}+13c-15=20\left(c-\frac{3}{5}\right)\left(c+\frac{5}{4}\right)$$

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

$$20c^{2}+13c-15=20\times \frac{5c-3}{5}\left(c+\frac{5}{4}\right)$$

Add \frac{5}{4} to c by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$20c^{2}+13c-15=20\times \frac{5c-3}{5}\times \frac{4c+5}{4}$$

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

$$20c^{2}+13c-15=20\times \frac{\left(5c-3\right)\left(4c+5\right)}{5\times 4}$$

Multiply 5 times 4.

$$20c^{2}+13c-15=20\times \frac{\left(5c-3\right)\left(4c+5\right)}{20}$$

Cancel out 20, the greatest common factor in 20 and 20.

$$20c^{2}+13c-15=\left(5c-3\right)\left(4c+5\right)$$

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 20

$$x ^ 2 +\frac{13}{20}x -\frac{3}{4} = 0$$

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

$$r + s = -\frac{13}{20} rs = -\frac{3}{4}$$

Two numbers r and s sum up to -\frac{13}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{13}{20} = -\frac{13}{40}. 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>

$$r = -\frac{13}{40} - u s = -\frac{13}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{4}

$$(-\frac{13}{40} - u) (-\frac{13}{40} + u) = -\frac{3}{4}$$

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

$$\frac{169}{1600} - u^2 = -\frac{3}{4}$$

Simplify the expression by subtracting \frac{169}{1600} on both sides

$$-u^2 = -\frac{3}{4}-\frac{169}{1600} = -\frac{1369}{1600}$$

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

$$u^2 = \frac{1369}{1600} u = \pm\sqrt{\frac{1369}{1600}} = \pm \frac{37}{40}$$

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

$$r =-\frac{13}{40} - \frac{37}{40} = -1.250 s = -\frac{13}{40} + \frac{37}{40} = 0.600$$

Evaluate

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