Solve 5y^2+31y+6

$$5 y ^ { 2 } + 31 y + 6$$

Factor

$$\left(y+6\right)\left(5y+1\right)$$

Solution Steps

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

$$a+b=31 ab=5\times 6=30$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 31.

$$a=1 b=30$$

Rewrite 5y^{2}+31y+6 as \left(5y^{2}+y\right)+\left(30y+6\right).

$$\left(5y^{2}+y\right)+\left(30y+6\right)$$

Factor out y in the first and 6 in the second group.

$$y\left(5y+1\right)+6\left(5y+1\right)$$

Factor out common term 5y+1 by using distributive property.

$$\left(5y+1\right)\left(y+6\right)$$

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.

$$5y^{2}+31y+6=0$$

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.

$$y=\frac{-31±\sqrt{31^{2}-4\times 5\times 6}}{2\times 5}$$

Square 31.

$$y=\frac{-31±\sqrt{961-4\times 5\times 6}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-31±\sqrt{961-20\times 6}}{2\times 5}$$

Multiply -20 times 6.

$$y=\frac{-31±\sqrt{961-120}}{2\times 5}$$

Add 961 to -120.

$$y=\frac{-31±\sqrt{841}}{2\times 5}$$

Take the square root of 841.

$$y=\frac{-31±29}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{-31±29}{10}$$

Now solve the equation y=\frac{-31±29}{10} when ± is plus. Add -31 to 29.

$$y=-\frac{2}{10}$$

Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.

$$y=-\frac{1}{5}$$

Now solve the equation y=\frac{-31±29}{10} when ± is minus. Subtract 29 from -31.

$$y=-\frac{60}{10}$$

Divide -60 by 10.

$$y=-6$$

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 -6 for x_{2}.

$$5y^{2}+31y+6=5\left(y-\left(-\frac{1}{5}\right)\right)\left(y-\left(-6\right)\right)$$

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

$$5y^{2}+31y+6=5\left(y+\frac{1}{5}\right)\left(y+6\right)$$

Add \frac{1}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$5y^{2}+31y+6=5\times \frac{5y+1}{5}\left(y+6\right)$$

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

$$5y^{2}+31y+6=\left(5y+1\right)\left(y+6\right)$$

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 5

$$x ^ 2 +\frac{31}{5}x +\frac{6}{5} = 0$$

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

$$r + s = -\frac{31}{5} rs = \frac{6}{5}$$

Two numbers r and s sum up to -\frac{31}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{31}{5} = -\frac{31}{10}. 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>

$$r = -\frac{31}{10} - u s = -\frac{31}{10} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{6}{5}

$$(-\frac{31}{10} - u) (-\frac{31}{10} + u) = \frac{6}{5}$$

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

$$\frac{961}{100} - u^2 = \frac{6}{5}$$

Simplify the expression by subtracting \frac{961}{100} on both sides

$$-u^2 = \frac{6}{5}-\frac{961}{100} = -\frac{841}{100}$$

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

$$u^2 = \frac{841}{100} u = \pm\sqrt{\frac{841}{100}} = \pm \frac{29}{10}$$

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

$$r =-\frac{31}{10} - \frac{29}{10} = -6 s = -\frac{31}{10} + \frac{29}{10} = -0.200$$

Evaluate

$$\left(y+6\right)\left(5y+1\right)$$