Solve y^2+13y+30

$${ y }^{ 2 } +13y+30$$

Factor

$$\left(y+3\right)\left(y+10\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as y^{2}+ay+by+30. To find a and b, set up a system to be solved.

$$a+b=13 ab=1\times 30=30$$

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 13.

$$a=3 b=10$$

Rewrite y^{2}+13y+30 as \left(y^{2}+3y\right)+\left(10y+30\right).

$$\left(y^{2}+3y\right)+\left(10y+30\right)$$

Factor out y in the first and 10 in the second group.

$$y\left(y+3\right)+10\left(y+3\right)$$

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

$$\left(y+3\right)\left(y+10\right)$$

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.

$$y^{2}+13y+30=0$$

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.

$$y=\frac{-13±\sqrt{13^{2}-4\times 30}}{2}$$

Square 13.

$$y=\frac{-13±\sqrt{169-4\times 30}}{2}$$

Multiply -4 times 30.

$$y=\frac{-13±\sqrt{169-120}}{2}$$

Add 169 to -120.

$$y=\frac{-13±\sqrt{49}}{2}$$

Take the square root of 49.

$$y=\frac{-13±7}{2}$$

Now solve the equation y=\frac{-13±7}{2} when ± is plus. Add -13 to 7.

$$y=-\frac{6}{2}$$

Divide -6 by 2.

$$y=-3$$

Now solve the equation y=\frac{-13±7}{2} when ± is minus. Subtract 7 from -13.

$$y=-\frac{20}{2}$$

Divide -20 by 2.

$$y=-10$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -3 for x_{1} and -10 for x_{2}.

$$y^{2}+13y+30=\left(y-\left(-3\right)\right)\left(y-\left(-10\right)\right)$$

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

$$y^{2}+13y+30=\left(y+3\right)\left(y+10\right)$$

Evaluate

$$\left(y+3\right)\left(y+10\right)$$