Solve 27y+10y^2+5

$$27 y + 10 y ^ { 2 } + 5$$

Factor

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

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$10y^{2}+27y+5$$

Factor the expression by grouping. First, the expression needs to be rewritten as 10y^{2}+ay+by+5. To find a and b, set up a system to be solved.

$$a+b=27 ab=10\times 5=50$$

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 50.

$$1,50 2,25 5,10$$

Calculate the sum for each pair.

$$1+50=51 2+25=27 5+10=15$$

The solution is the pair that gives sum 27.

$$a=2 b=25$$

Rewrite 10y^{2}+27y+5 as \left(10y^{2}+2y\right)+\left(25y+5\right).

$$\left(10y^{2}+2y\right)+\left(25y+5\right)$$

Factor out 2y in the first and 5 in the second group.

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

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

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

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.

$$10y^{2}+27y+5=0$$

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.

$$y=\frac{-27±\sqrt{27^{2}-4\times 10\times 5}}{2\times 10}$$

Square 27.

$$y=\frac{-27±\sqrt{729-4\times 10\times 5}}{2\times 10}$$

Multiply -4 times 10.

$$y=\frac{-27±\sqrt{729-40\times 5}}{2\times 10}$$

Multiply -40 times 5.

$$y=\frac{-27±\sqrt{729-200}}{2\times 10}$$

Add 729 to -200.

$$y=\frac{-27±\sqrt{529}}{2\times 10}$$

Take the square root of 529.

$$y=\frac{-27±23}{2\times 10}$$

Multiply 2 times 10.

$$y=\frac{-27±23}{20}$$

Now solve the equation y=\frac{-27±23}{20} when ± is plus. Add -27 to 23.

$$y=-\frac{4}{20}$$

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

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

Now solve the equation y=\frac{-27±23}{20} when ± is minus. Subtract 23 from -27.

$$y=-\frac{50}{20}$$

Reduce the fraction \frac{-50}{20} to lowest terms by extracting and canceling out 10.

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

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 -\frac{5}{2} for x_{2}.

$$10y^{2}+27y+5=10\left(y-\left(-\frac{1}{5}\right)\right)\left(y-\left(-\frac{5}{2}\right)\right)$$

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

$$10y^{2}+27y+5=10\left(y+\frac{1}{5}\right)\left(y+\frac{5}{2}\right)$$

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

$$10y^{2}+27y+5=10\times \frac{5y+1}{5}\left(y+\frac{5}{2}\right)$$

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

$$10y^{2}+27y+5=10\times \frac{5y+1}{5}\times \frac{2y+5}{2}$$

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

$$10y^{2}+27y+5=10\times \frac{\left(5y+1\right)\left(2y+5\right)}{5\times 2}$$

Multiply 5 times 2.

$$10y^{2}+27y+5=10\times \frac{\left(5y+1\right)\left(2y+5\right)}{10}$$

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

$$10y^{2}+27y+5=\left(5y+1\right)\left(2y+5\right)$$

Evaluate

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