Solve 15y^2-18y-24

$$15 y ^ { 2 } - 18 y - 24$$

Factor

$$3\left(y-2\right)\left(5y+4\right)$$

Solution Steps

Factor out 3.

$$3\left(5y^{2}-6y-8\right)$$

Consider 5y^{2}-6y-8. Factor the expression by grouping. First, the expression needs to be rewritten as 5y^{2}+ay+by-8. To find a and b, set up a system to be solved.

$$a+b=-6 ab=5\left(-8\right)=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -6.

$$a=-10 b=4$$

Rewrite 5y^{2}-6y-8 as \left(5y^{2}-10y\right)+\left(4y-8\right).

$$\left(5y^{2}-10y\right)+\left(4y-8\right)$$

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

$$5y\left(y-2\right)+4\left(y-2\right)$$

Factor out common term y-2 by using distributive property.

$$\left(y-2\right)\left(5y+4\right)$$

Rewrite the complete factored expression.

$$3\left(y-2\right)\left(5y+4\right)$$

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.

$$15y^{2}-18y-24=0$$

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.

$$y=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 15\left(-24\right)}}{2\times 15}$$

Square -18.

$$y=\frac{-\left(-18\right)±\sqrt{324-4\times 15\left(-24\right)}}{2\times 15}$$

Multiply -4 times 15.

$$y=\frac{-\left(-18\right)±\sqrt{324-60\left(-24\right)}}{2\times 15}$$

Multiply -60 times -24.

$$y=\frac{-\left(-18\right)±\sqrt{324+1440}}{2\times 15}$$

Add 324 to 1440.

$$y=\frac{-\left(-18\right)±\sqrt{1764}}{2\times 15}$$

Take the square root of 1764.

$$y=\frac{-\left(-18\right)±42}{2\times 15}$$

The opposite of -18 is 18.

$$y=\frac{18±42}{2\times 15}$$

Multiply 2 times 15.

$$y=\frac{18±42}{30}$$

Now solve the equation y=\frac{18±42}{30} when ± is plus. Add 18 to 42.

$$y=\frac{60}{30}$$

Divide 60 by 30.

$$y=2$$

Now solve the equation y=\frac{18±42}{30} when ± is minus. Subtract 42 from 18.

$$y=-\frac{24}{30}$$

Reduce the fraction \frac{-24}{30} to lowest terms by extracting and canceling out 6.

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

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

$$15y^{2}-18y-24=15\left(y-2\right)\left(y-\left(-\frac{4}{5}\right)\right)$$

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

$$15y^{2}-18y-24=15\left(y-2\right)\left(y+\frac{4}{5}\right)$$

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

$$15y^{2}-18y-24=15\left(y-2\right)\times \frac{5y+4}{5}$$

Cancel out 5, the greatest common factor in 15 and 5.

$$15y^{2}-18y-24=3\left(y-2\right)\left(5y+4\right)$$

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 15

$$x ^ 2 -\frac{6}{5}x -\frac{8}{5} = 0$$

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

$$r + s = \frac{6}{5} rs = -\frac{8}{5}$$

Two numbers r and s sum up to \frac{6}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{6}{5} = \frac{3}{5}. 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>

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

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{8}{5}

$$(\frac{3}{5} - u) (\frac{3}{5} + u) = -\frac{8}{5}$$

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

$$\frac{9}{25} - u^2 = -\frac{8}{5}$$

Simplify the expression by subtracting \frac{9}{25} on both sides

$$-u^2 = -\frac{8}{5}-\frac{9}{25} = -\frac{49}{25}$$

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

$$u^2 = \frac{49}{25} u = \pm\sqrt{\frac{49}{25}} = \pm \frac{7}{5}$$

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

$$r =\frac{3}{5} - \frac{7}{5} = -0.800 s = \frac{3}{5} + \frac{7}{5} = 2$$

Evaluate

$$3\left(y-2\right)\left(5y+4\right)$$