Solve 15y^2+19y+6

$$15 y ^ { 2 } + 19 y + 6$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 15y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=19 ab=15\times 6=90$$

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 90.

$$1,90 2,45 3,30 5,18 6,15 9,10$$

Calculate the sum for each pair.

$$1+90=91 2+45=47 3+30=33 5+18=23 6+15=21 9+10=19$$

The solution is the pair that gives sum 19.

$$a=9 b=10$$

Rewrite 15y^{2}+19y+6 as \left(15y^{2}+9y\right)+\left(10y+6\right).

$$\left(15y^{2}+9y\right)+\left(10y+6\right)$$

Factor out 3y in the first and 2 in the second group.

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

Factor out common term 5y+3 by using distributive property.

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

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.

$$15y^{2}+19y+6=0$$

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.

$$y=\frac{-19±\sqrt{19^{2}-4\times 15\times 6}}{2\times 15}$$

Square 19.

$$y=\frac{-19±\sqrt{361-4\times 15\times 6}}{2\times 15}$$

Multiply -4 times 15.

$$y=\frac{-19±\sqrt{361-60\times 6}}{2\times 15}$$

Multiply -60 times 6.

$$y=\frac{-19±\sqrt{361-360}}{2\times 15}$$

Add 361 to -360.

$$y=\frac{-19±\sqrt{1}}{2\times 15}$$

Take the square root of 1.

$$y=\frac{-19±1}{2\times 15}$$

Multiply 2 times 15.

$$y=\frac{-19±1}{30}$$

Now solve the equation y=\frac{-19±1}{30} when ± is plus. Add -19 to 1.

$$y=-\frac{18}{30}$$

Reduce the fraction \frac{-18}{30} to lowest terms by extracting and canceling out 6.

$$y=-\frac{3}{5}$$

Now solve the equation y=\frac{-19±1}{30} when ± is minus. Subtract 1 from -19.

$$y=-\frac{20}{30}$$

Reduce the fraction \frac{-20}{30} to lowest terms by extracting and canceling out 10.

$$y=-\frac{2}{3}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{3}{5} for x_{1} and -\frac{2}{3} for x_{2}.

$$15y^{2}+19y+6=15\left(y-\left(-\frac{3}{5}\right)\right)\left(y-\left(-\frac{2}{3}\right)\right)$$

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

$$15y^{2}+19y+6=15\left(y+\frac{3}{5}\right)\left(y+\frac{2}{3}\right)$$

Add \frac{3}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15y^{2}+19y+6=15\times \frac{5y+3}{5}\left(y+\frac{2}{3}\right)$$

Add \frac{2}{3} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$15y^{2}+19y+6=15\times \frac{5y+3}{5}\times \frac{3y+2}{3}$$

Multiply \frac{5y+3}{5} times \frac{3y+2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$15y^{2}+19y+6=15\times \frac{\left(5y+3\right)\left(3y+2\right)}{5\times 3}$$

Multiply 5 times 3.

$$15y^{2}+19y+6=15\times \frac{\left(5y+3\right)\left(3y+2\right)}{15}$$

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

$$15y^{2}+19y+6=\left(5y+3\right)\left(3y+2\right)$$

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 15

$$x ^ 2 +\frac{19}{15}x +\frac{2}{5} = 0$$

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

$$r + s = -\frac{19}{15} rs = \frac{2}{5}$$

Two numbers r and s sum up to -\frac{19}{15} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{15} = -\frac{19}{30}. 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>

$$r = -\frac{19}{30} - u s = -\frac{19}{30} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{5}

$$(-\frac{19}{30} - u) (-\frac{19}{30} + u) = \frac{2}{5}$$

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

$$\frac{361}{900} - u^2 = \frac{2}{5}$$

Simplify the expression by subtracting \frac{361}{900} on both sides

$$-u^2 = \frac{2}{5}-\frac{361}{900} = -\frac{1}{900}$$

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

$$u^2 = \frac{1}{900} u = \pm\sqrt{\frac{1}{900}} = \pm \frac{1}{30}$$

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

$$r =-\frac{19}{30} - \frac{1}{30} = -0.667 s = -\frac{19}{30} + \frac{1}{30} = -0.600$$

Evaluate

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