Question

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

Answer 1

`2x^2 + 13x + 6 = 2 (x + 1/2)(x+6)`

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, `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!

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

`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`

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

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)`