# Is it possible to factor #y=2x^2 + 13x + 6 #? If so, what are the factors?

##### 1 Answer

#### Explanation:

If it is possible to factor

#y = 2x^2 + 13x + 6# ,

then your equation can be written as

#y = a (x + r)(x + s)#

Here,

#y = 2 (x^2 + 13/2 x + 3)#

# = 2 (x + r)(x+s)#

One way to find such numbers

To do so, you need to set

This can be done e.g. with a quadratic formula. However, let me show you one of my favorite methods to find solutions of a quadratic equation: completion of the circle.

*If you are not interested and would rather do it with the quadratic formula, you can skip the next part!*

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# x^2 + 13/2 x + 3 = 0#

... compute

# x^2 + 13/2x = -3#

Now, on the left side, we would like to have something like

#a^2 + 2ab + b^2 = (a+b)^2#

We already have

Now, to create an

# x^2 + 13/2x + (13/4)^2= -3 + (13/4)^2#

... apply

# (x +13/4)^2 = 121/16#

Now, you can draw the root on the left side, but be careful: when doing so, you are creating two solutions since e.g. for

Thus, we get

#x + 13/4 = sqrt(121/16) " or " x + 13/4 = - sqrt(121/16)#

#x = - 1/2 " or " x = 6#

======================

Thus, the equation

#2x^2 + 13x + 6 = 0#

#2 (x^2 + 13/2 x + 3) = 0#

has two solutions:

# x = - 1/2 " or " x = 6#

Thus, you can factorize using negative values of the two solutions:

#2x^2 + 13x + 6 = 2 (x - ( - 1/2) )(x - 6) = 2 (x + 1/2)(x+6)#