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

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, #a# is the coefficient of the #x^2# term, so #a= 2#.

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

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

One way to find such numbers #r# and #s# is computing the roots (zeros) of the polynomial.

To do so, you need to set #x^2 + 13/2x + 3 = 0# and search for possible solutions.

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 #-3# on both sides of the equation...

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

Now, on the left side, we would like to have something like #a^2 + 2ab + b^2# so that we are able to apply the formula

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

We already have #a^2 = x^2#, so our #a = x#, and we also have #2ab = 13/2 x#. Thus, we can conclude that #b = 13/4#.

Now, to create an #a^2 + 2ab + b^2# expression on the left side, we need to add our #b^2# term, namely #(13/4)^2#. However, since we don't want to jeopardize the equality, we need to add #(13/4)^2# on the right side of the equation as well:

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

... apply #a^2 + 2ab + b^2 = (a+b)^2# on the left side and calculate the right side...

# (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 #x^2 = 25#, both #x = 5# and #x = -5# are solutions.

Thus, we get

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

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

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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)#