Solve 60s^2+33s-9

$$60 s ^ { 2 } + 33 s - 9$$

Factor

$$3\left(5s-1\right)\left(4s+3\right)$$

Solution Steps

Factor out 3.

$$3\left(20s^{2}+11s-3\right)$$

Consider 20s^{2}+11s-3. Factor the expression by grouping. First, the expression needs to be rewritten as 20s^{2}+as+bs-3. To find a and b, set up a system to be solved.

$$a+b=11 ab=20\left(-3\right)=-60$$

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 -60.

$$-1,60 -2,30 -3,20 -4,15 -5,12 -6,10$$

Calculate the sum for each pair.

$$-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4$$

The solution is the pair that gives sum 11.

$$a=-4 b=15$$

Rewrite 20s^{2}+11s-3 as \left(20s^{2}-4s\right)+\left(15s-3\right).

$$\left(20s^{2}-4s\right)+\left(15s-3\right)$$

Factor out 4s in the first and 3 in the second group.

$$4s\left(5s-1\right)+3\left(5s-1\right)$$

Factor out common term 5s-1 by using distributive property.

$$\left(5s-1\right)\left(4s+3\right)$$

Rewrite the complete factored expression.

$$3\left(5s-1\right)\left(4s+3\right)$$

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.

$$60s^{2}+33s-9=0$$

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.

$$s=\frac{-33±\sqrt{33^{2}-4\times 60\left(-9\right)}}{2\times 60}$$

Square 33.

$$s=\frac{-33±\sqrt{1089-4\times 60\left(-9\right)}}{2\times 60}$$

Multiply -4 times 60.

$$s=\frac{-33±\sqrt{1089-240\left(-9\right)}}{2\times 60}$$

Multiply -240 times -9.

$$s=\frac{-33±\sqrt{1089+2160}}{2\times 60}$$

Add 1089 to 2160.

$$s=\frac{-33±\sqrt{3249}}{2\times 60}$$

Take the square root of 3249.

$$s=\frac{-33±57}{2\times 60}$$

Multiply 2 times 60.

$$s=\frac{-33±57}{120}$$

Now solve the equation s=\frac{-33±57}{120} when ± is plus. Add -33 to 57.

$$s=\frac{24}{120}$$

Reduce the fraction \frac{24}{120} to lowest terms by extracting and canceling out 24.

$$s=\frac{1}{5}$$

Now solve the equation s=\frac{-33±57}{120} when ± is minus. Subtract 57 from -33.

$$s=-\frac{90}{120}$$

Reduce the fraction \frac{-90}{120} to lowest terms by extracting and canceling out 30.

$$s=-\frac{3}{4}$$

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

$$60s^{2}+33s-9=60\left(s-\frac{1}{5}\right)\left(s-\left(-\frac{3}{4}\right)\right)$$

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

$$60s^{2}+33s-9=60\left(s-\frac{1}{5}\right)\left(s+\frac{3}{4}\right)$$

Subtract \frac{1}{5} from s by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$60s^{2}+33s-9=60\times \frac{5s-1}{5}\left(s+\frac{3}{4}\right)$$

Add \frac{3}{4} to s by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$60s^{2}+33s-9=60\times \frac{5s-1}{5}\times \frac{4s+3}{4}$$

Multiply \frac{5s-1}{5} times \frac{4s+3}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$60s^{2}+33s-9=60\times \frac{\left(5s-1\right)\left(4s+3\right)}{5\times 4}$$

Multiply 5 times 4.

$$60s^{2}+33s-9=60\times \frac{\left(5s-1\right)\left(4s+3\right)}{20}$$

Cancel out 20, the greatest common factor in 60 and 20.

$$60s^{2}+33s-9=3\left(5s-1\right)\left(4s+3\right)$$

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 60

$$x ^ 2 +\frac{11}{20}x -\frac{3}{20} = 0$$

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

$$r + s = -\frac{11}{20} rs = -\frac{3}{20}$$

Two numbers r and s sum up to -\frac{11}{20} exactly when the average of the two numbers is \frac{1}{2}*-\frac{11}{20} = -\frac{11}{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>

$$r = -\frac{11}{40} - u s = -\frac{11}{40} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{20}

$$(-\frac{11}{40} - u) (-\frac{11}{40} + u) = -\frac{3}{20}$$

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

$$\frac{121}{1600} - u^2 = -\frac{3}{20}$$

Simplify the expression by subtracting \frac{121}{1600} on both sides

$$-u^2 = -\frac{3}{20}-\frac{121}{1600} = -\frac{361}{1600}$$

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

$$u^2 = \frac{361}{1600} u = \pm\sqrt{\frac{361}{1600}} = \pm \frac{19}{40}$$

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

$$r =-\frac{11}{40} - \frac{19}{40} = -0.750 s = -\frac{11}{40} + \frac{19}{40} = 0.200$$

Evaluate

$$60s^{2}+33s-9$$