Solve 10y^2+19y+6

$$10 { y }^{ 2 } +19y+6$$

Factor

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

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10y^{2}+ay+by+6. To find a and b, set up a system to be solved.

$$a+b=19 ab=10\times 6=60$$

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 60.

$$1,60 2,30 3,20 4,15 5,12 6,10$$

Calculate the sum for each pair.

$$1+60=61 2+30=32 3+20=23 4+15=19 5+12=17 6+10=16$$

The solution is the pair that gives sum 19.

$$a=4 b=15$$

Rewrite 10y^{2}+19y+6 as \left(10y^{2}+4y\right)+\left(15y+6\right).

$$\left(10y^{2}+4y\right)+\left(15y+6\right)$$

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

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

Factor out common term 5y+2 by using distributive property.

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

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.

$$10y^{2}+19y+6=0$$

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.

$$y=\frac{-19±\sqrt{19^{2}-4\times 10\times 6}}{2\times 10}$$

Square 19.

$$y=\frac{-19±\sqrt{361-4\times 10\times 6}}{2\times 10}$$

Multiply -4 times 10.

$$y=\frac{-19±\sqrt{361-40\times 6}}{2\times 10}$$

Multiply -40 times 6.

$$y=\frac{-19±\sqrt{361-240}}{2\times 10}$$

Add 361 to -240.

$$y=\frac{-19±\sqrt{121}}{2\times 10}$$

Take the square root of 121.

$$y=\frac{-19±11}{2\times 10}$$

Multiply 2 times 10.

$$y=\frac{-19±11}{20}$$

Now solve the equation y=\frac{-19±11}{20} when ± is plus. Add -19 to 11.

$$y=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

$$y=-\frac{2}{5}$$

Now solve the equation y=\frac{-19±11}{20} when ± is minus. Subtract 11 from -19.

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

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

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

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

$$10y^{2}+19y+6=10\left(y-\left(-\frac{2}{5}\right)\right)\left(y-\left(-\frac{3}{2}\right)\right)$$

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

$$10y^{2}+19y+6=10\left(y+\frac{2}{5}\right)\left(y+\frac{3}{2}\right)$$

Add \frac{2}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10y^{2}+19y+6=10\times \frac{5y+2}{5}\left(y+\frac{3}{2}\right)$$

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

$$10y^{2}+19y+6=10\times \frac{5y+2}{5}\times \frac{2y+3}{2}$$

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

$$10y^{2}+19y+6=10\times \frac{\left(5y+2\right)\left(2y+3\right)}{5\times 2}$$

Multiply 5 times 2.

$$10y^{2}+19y+6=10\times \frac{\left(5y+2\right)\left(2y+3\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10y^{2}+19y+6=\left(5y+2\right)\left(2y+3\right)$$

Evaluate

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