Solve 14y^2-3y-5

$$14 y ^ { 2 } - 3 y - 5$$

Factor

$$\left(7y-5\right)\left(2y+1\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 14y^{2}+ay+by-5. To find a and b, set up a system to be solved.

$$a+b=-3 ab=14\left(-5\right)=-70$$

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 -70.

$$1,-70 2,-35 5,-14 7,-10$$

Calculate the sum for each pair.

$$1-70=-69 2-35=-33 5-14=-9 7-10=-3$$

The solution is the pair that gives sum -3.

$$a=-10 b=7$$

Rewrite 14y^{2}-3y-5 as \left(14y^{2}-10y\right)+\left(7y-5\right).

$$\left(14y^{2}-10y\right)+\left(7y-5\right)$$

Factor out 2y in 14y^{2}-10y.

$$2y\left(7y-5\right)+7y-5$$

Factor out common term 7y-5 by using distributive property.

$$\left(7y-5\right)\left(2y+1\right)$$

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.

$$14y^{2}-3y-5=0$$

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.

$$y=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 14\left(-5\right)}}{2\times 14}$$

Square -3.

$$y=\frac{-\left(-3\right)±\sqrt{9-4\times 14\left(-5\right)}}{2\times 14}$$

Multiply -4 times 14.

$$y=\frac{-\left(-3\right)±\sqrt{9-56\left(-5\right)}}{2\times 14}$$

Multiply -56 times -5.

$$y=\frac{-\left(-3\right)±\sqrt{9+280}}{2\times 14}$$

Add 9 to 280.

$$y=\frac{-\left(-3\right)±\sqrt{289}}{2\times 14}$$

Take the square root of 289.

$$y=\frac{-\left(-3\right)±17}{2\times 14}$$

The opposite of -3 is 3.

$$y=\frac{3±17}{2\times 14}$$

Multiply 2 times 14.

$$y=\frac{3±17}{28}$$

Now solve the equation y=\frac{3±17}{28} when ± is plus. Add 3 to 17.

$$y=\frac{20}{28}$$

Reduce the fraction \frac{20}{28} to lowest terms by extracting and canceling out 4.

$$y=\frac{5}{7}$$

Now solve the equation y=\frac{3±17}{28} when ± is minus. Subtract 17 from 3.

$$y=-\frac{14}{28}$$

Reduce the fraction \frac{-14}{28} to lowest terms by extracting and canceling out 14.

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

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{7} for x_{1} and -\frac{1}{2} for x_{2}.

$$14y^{2}-3y-5=14\left(y-\frac{5}{7}\right)\left(y-\left(-\frac{1}{2}\right)\right)$$

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

$$14y^{2}-3y-5=14\left(y-\frac{5}{7}\right)\left(y+\frac{1}{2}\right)$$

Subtract \frac{5}{7} from y by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$14y^{2}-3y-5=14\times \frac{7y-5}{7}\left(y+\frac{1}{2}\right)$$

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

$$14y^{2}-3y-5=14\times \frac{7y-5}{7}\times \frac{2y+1}{2}$$

Multiply \frac{7y-5}{7} times \frac{2y+1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$14y^{2}-3y-5=14\times \frac{\left(7y-5\right)\left(2y+1\right)}{7\times 2}$$

Multiply 7 times 2.

$$14y^{2}-3y-5=14\times \frac{\left(7y-5\right)\left(2y+1\right)}{14}$$

Cancel out 14, the greatest common factor in 14 and 14.

$$14y^{2}-3y-5=\left(7y-5\right)\left(2y+1\right)$$

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 14

$$x ^ 2 -\frac{3}{14}x -\frac{5}{14} = 0$$

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

$$r + s = \frac{3}{14} rs = -\frac{5}{14}$$

Two numbers r and s sum up to \frac{3}{14} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{14} = \frac{3}{28}. 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>

$$r = \frac{3}{28} - u s = \frac{3}{28} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{5}{14}

$$(\frac{3}{28} - u) (\frac{3}{28} + u) = -\frac{5}{14}$$

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

$$\frac{9}{784} - u^2 = -\frac{5}{14}$$

Simplify the expression by subtracting \frac{9}{784} on both sides

$$-u^2 = -\frac{5}{14}-\frac{9}{784} = -\frac{289}{784}$$

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

$$u^2 = \frac{289}{784} u = \pm\sqrt{\frac{289}{784}} = \pm \frac{17}{28}$$

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

$$r =\frac{3}{28} - \frac{17}{28} = -0.500 s = \frac{3}{28} + \frac{17}{28} = 0.714$$

Evaluate

$$\left(7y-5\right)\left(2y+1\right)$$