Solve 40y^2-10y-15

$$40 y ^ { 2 } - 10 y - 15$$

Factor

$$5\left(4y-3\right)\left(2y+1\right)$$

Solution Steps

Factor out 5.

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

Consider 8y^{2}-2y-3. Factor the expression by grouping. First, the expression needs to be rewritten as 8y^{2}+ay+by-3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=8\left(-3\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -2.

$$a=-6 b=4$$

Rewrite 8y^{2}-2y-3 as \left(8y^{2}-6y\right)+\left(4y-3\right).

$$\left(8y^{2}-6y\right)+\left(4y-3\right)$$

Factor out 2y in 8y^{2}-6y.

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

Factor out common term 4y-3 by using distributive property.

$$\left(4y-3\right)\left(2y+1\right)$$

Rewrite the complete factored expression.

$$5\left(4y-3\right)\left(2y+1\right)$$

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.

$$40y^{2}-10y-15=0$$

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.

$$y=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 40\left(-15\right)}}{2\times 40}$$

Square -10.

$$y=\frac{-\left(-10\right)±\sqrt{100-4\times 40\left(-15\right)}}{2\times 40}$$

Multiply -4 times 40.

$$y=\frac{-\left(-10\right)±\sqrt{100-160\left(-15\right)}}{2\times 40}$$

Multiply -160 times -15.

$$y=\frac{-\left(-10\right)±\sqrt{100+2400}}{2\times 40}$$

Add 100 to 2400.

$$y=\frac{-\left(-10\right)±\sqrt{2500}}{2\times 40}$$

Take the square root of 2500.

$$y=\frac{-\left(-10\right)±50}{2\times 40}$$

The opposite of -10 is 10.

$$y=\frac{10±50}{2\times 40}$$

Multiply 2 times 40.

$$y=\frac{10±50}{80}$$

Now solve the equation y=\frac{10±50}{80} when ± is plus. Add 10 to 50.

$$y=\frac{60}{80}$$

Reduce the fraction \frac{60}{80} to lowest terms by extracting and canceling out 20.

$$y=\frac{3}{4}$$

Now solve the equation y=\frac{10±50}{80} when ± is minus. Subtract 50 from 10.

$$y=-\frac{40}{80}$$

Reduce the fraction \frac{-40}{80} to lowest terms by extracting and canceling out 40.

$$y=-\frac{1}{2}$$

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

$$40y^{2}-10y-15=40\left(y-\frac{3}{4}\right)\left(y-\left(-\frac{1}{2}\right)\right)$$

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

$$40y^{2}-10y-15=40\left(y-\frac{3}{4}\right)\left(y+\frac{1}{2}\right)$$

Subtract \frac{3}{4} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$40y^{2}-10y-15=40\times \frac{4y-3}{4}\left(y+\frac{1}{2}\right)$$

Add \frac{1}{2} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$40y^{2}-10y-15=40\times \frac{4y-3}{4}\times \frac{2y+1}{2}$$

Multiply \frac{4y-3}{4} times \frac{2y+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$40y^{2}-10y-15=40\times \frac{\left(4y-3\right)\left(2y+1\right)}{4\times 2}$$

Multiply 4 times 2.

$$40y^{2}-10y-15=40\times \frac{\left(4y-3\right)\left(2y+1\right)}{8}$$

Cancel out 8, the greatest common factor in 40 and 8.

$$40y^{2}-10y-15=5\left(4y-3\right)\left(2y+1\right)$$

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 40

$$x ^ 2 -\frac{1}{4}x -\frac{3}{8} = 0$$

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

$$r + s = \frac{1}{4} rs = -\frac{3}{8}$$

Two numbers r and s sum up to \frac{1}{4} exactly when the average of the two numbers is \frac{1}{2}*\frac{1}{4} = \frac{1}{8}. 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>

$$r = \frac{1}{8} - u s = \frac{1}{8} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{3}{8}

$$(\frac{1}{8} - u) (\frac{1}{8} + u) = -\frac{3}{8}$$

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

$$\frac{1}{64} - u^2 = -\frac{3}{8}$$

Simplify the expression by subtracting \frac{1}{64} on both sides

$$-u^2 = -\frac{3}{8}-\frac{1}{64} = -\frac{25}{64}$$

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

$$u^2 = \frac{25}{64} u = \pm\sqrt{\frac{25}{64}} = \pm \frac{5}{8}$$

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

$$r =\frac{1}{8} - \frac{5}{8} = -0.500 s = \frac{1}{8} + \frac{5}{8} = 0.750$$

Evaluate

$$40y^{2}-10y-15$$