Solve 7y^2-19y-6

$$7 y ^ { 2 } - 19 y - 6$$

Factor

$$\left(y-3\right)\left(7y+2\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 7y^{2}+ay+by-6. To find a and b, set up a system to be solved.

$$a+b=-19 ab=7\left(-6\right)=-42$$

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 -42.

$$1,-42 2,-21 3,-14 6,-7$$

Calculate the sum for each pair.

$$1-42=-41 2-21=-19 3-14=-11 6-7=-1$$

The solution is the pair that gives sum -19.

$$a=-21 b=2$$

Rewrite 7y^{2}-19y-6 as \left(7y^{2}-21y\right)+\left(2y-6\right).

$$\left(7y^{2}-21y\right)+\left(2y-6\right)$$

Factor out 7y in the first and 2 in the second group.

$$7y\left(y-3\right)+2\left(y-3\right)$$

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

$$\left(y-3\right)\left(7y+2\right)$$

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.

$$7y^{2}-19y-6=0$$

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.

$$y=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 7\left(-6\right)}}{2\times 7}$$

Square -19.

$$y=\frac{-\left(-19\right)±\sqrt{361-4\times 7\left(-6\right)}}{2\times 7}$$

Multiply -4 times 7.

$$y=\frac{-\left(-19\right)±\sqrt{361-28\left(-6\right)}}{2\times 7}$$

Multiply -28 times -6.

$$y=\frac{-\left(-19\right)±\sqrt{361+168}}{2\times 7}$$

Add 361 to 168.

$$y=\frac{-\left(-19\right)±\sqrt{529}}{2\times 7}$$

Take the square root of 529.

$$y=\frac{-\left(-19\right)±23}{2\times 7}$$

The opposite of -19 is 19.

$$y=\frac{19±23}{2\times 7}$$

Multiply 2 times 7.

$$y=\frac{19±23}{14}$$

Now solve the equation y=\frac{19±23}{14} when ± is plus. Add 19 to 23.

$$y=\frac{42}{14}$$

Divide 42 by 14.

$$y=3$$

Now solve the equation y=\frac{19±23}{14} when ± is minus. Subtract 23 from 19.

$$y=-\frac{4}{14}$$

Reduce the fraction \frac{-4}{14} to lowest terms by extracting and canceling out 2.

$$y=-\frac{2}{7}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3 for x_{1} and -\frac{2}{7} for x_{2}.

$$7y^{2}-19y-6=7\left(y-3\right)\left(y-\left(-\frac{2}{7}\right)\right)$$

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

$$7y^{2}-19y-6=7\left(y-3\right)\left(y+\frac{2}{7}\right)$$

Add \frac{2}{7} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$7y^{2}-19y-6=7\left(y-3\right)\times \frac{7y+2}{7}$$

Cancel out 7, the greatest common factor in 7 and 7.

$$7y^{2}-19y-6=\left(y-3\right)\left(7y+2\right)$$

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 7

$$x ^ 2 -\frac{19}{7}x -\frac{6}{7} = 0$$

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

$$r + s = \frac{19}{7} rs = -\frac{6}{7}$$

Two numbers r and s sum up to \frac{19}{7} exactly when the average of the two numbers is \frac{1}{2}*\frac{19}{7} = \frac{19}{14}. 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>

$$r = \frac{19}{14} - u s = \frac{19}{14} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{6}{7}

$$(\frac{19}{14} - u) (\frac{19}{14} + u) = -\frac{6}{7}$$

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

$$\frac{361}{196} - u^2 = -\frac{6}{7}$$

Simplify the expression by subtracting \frac{361}{196} on both sides

$$-u^2 = -\frac{6}{7}-\frac{361}{196} = -\frac{529}{196}$$

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

$$u^2 = \frac{529}{196} u = \pm\sqrt{\frac{529}{196}} = \pm \frac{23}{14}$$

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

$$r =\frac{19}{14} - \frac{23}{14} = -0.286 s = \frac{19}{14} + \frac{23}{14} = 3$$

Evaluate

$$\left(y-3\right)\left(7y+2\right)$$