Solve 5y^2+50y+45

$$5 { y }^{ 2 } +50y+45$$

Factor

$$5\left(y+1\right)\left(y+9\right)$$

Solution Steps

Factor out 5.

$$5\left(y^{2}+10y+9\right)$$

Consider y^{2}+10y+9. Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+9. To find a and b, set up a system to be solved.

$$a+b=10 ab=1\times 9=9$$

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 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=1 b=9$$

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

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

Factor out y in the first and 9 in the second group.

$$y\left(y+1\right)+9\left(y+1\right)$$

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

$$\left(y+1\right)\left(y+9\right)$$

Rewrite the complete factored expression.

$$5\left(y+1\right)\left(y+9\right)$$

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.

$$5y^{2}+50y+45=0$$

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.

$$y=\frac{-50±\sqrt{50^{2}-4\times 5\times 45}}{2\times 5}$$

Square 50.

$$y=\frac{-50±\sqrt{2500-4\times 5\times 45}}{2\times 5}$$

Multiply -4 times 5.

$$y=\frac{-50±\sqrt{2500-20\times 45}}{2\times 5}$$

Multiply -20 times 45.

$$y=\frac{-50±\sqrt{2500-900}}{2\times 5}$$

Add 2500 to -900.

$$y=\frac{-50±\sqrt{1600}}{2\times 5}$$

Take the square root of 1600.

$$y=\frac{-50±40}{2\times 5}$$

Multiply 2 times 5.

$$y=\frac{-50±40}{10}$$

Now solve the equation y=\frac{-50±40}{10} when ± is plus. Add -50 to 40.

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

Divide -10 by 10.

$$y=-1$$

Now solve the equation y=\frac{-50±40}{10} when ± is minus. Subtract 40 from -50.

$$y=-\frac{90}{10}$$

Divide -90 by 10.

$$y=-9$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -1 for x_{1} and -9 for x_{2}.

$$5y^{2}+50y+45=5\left(y-\left(-1\right)\right)\left(y-\left(-9\right)\right)$$

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

$$5y^{2}+50y+45=5\left(y+1\right)\left(y+9\right)$$

Evaluate

$$5\left(y+1\right)\left(y+9\right)$$