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

Jan 22, 2016

$2 {x}^{2} + 13 x + 6 = 2 \left(x + \frac{1}{2}\right) \left(x + 6\right)$

#### Explanation:

If it is possible to factor

$y = 2 {x}^{2} + 13 x + 6$,

then your equation can be written as

$y = a \left(x + r\right) \left(x + s\right)$

Here, $a$ is the coefficient of the ${x}^{2}$ term, so $a = 2$.

$y = 2 \left({x}^{2} + \frac{13}{2} x + 3\right)$

$= 2 \left(x + r\right) \left(x + s\right)$

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

To do so, you need to set ${x}^{2} + \frac{13}{2} x + 3 = 0$ 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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${x}^{2} + \frac{13}{2} x + 3 = 0$

... compute $- 3$ on both sides of the equation...

${x}^{2} + \frac{13}{2} x = - 3$

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

${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

We already have ${a}^{2} = {x}^{2}$, so our $a = x$, and we also have $2 a b = \frac{13}{2} x$. Thus, we can conclude that $b = \frac{13}{4}$.

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

${x}^{2} + \frac{13}{2} x + {\left(\frac{13}{4}\right)}^{2} = - 3 + {\left(\frac{13}{4}\right)}^{2}$

... apply ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$ on the left side and calculate the right side...

${\left(x + \frac{13}{4}\right)}^{2} = \frac{121}{16}$

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

$x + \frac{13}{4} = \sqrt{\frac{121}{16}} \text{ or } x + \frac{13}{4} = - \sqrt{\frac{121}{16}}$

$x = - \frac{1}{2} \text{ or } x = 6$

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Thus, the equation

$2 {x}^{2} + 13 x + 6 = 0$

$2 \left({x}^{2} + \frac{13}{2} x + 3\right) = 0$

has two solutions:

$x = - \frac{1}{2} \text{ or } x = 6$

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

$2 {x}^{2} + 13 x + 6 = 2 \left(x - \left(- \frac{1}{2}\right)\right) \left(x - 6\right) = 2 \left(x + \frac{1}{2}\right) \left(x + 6\right)$