# Factor x^2 +12x +36?

Jul 15, 2018

See a solution process below:

#### Explanation:

Because the ${x}^{2}$ coefficient is $1$ we know the coefficient for the $x$ terms in the factor will also be $1$:

$\left(x\right) \left(x\right)$

Because the constant is a positive and the coefficient for the $x$ term is also a positive we know the sign for the constants in the factors will both be positive:

$\left(x +\right) \left(x +\right)$

Now we need to determine the factors which multiply to 36 and also add to 12:

$1 \times 36 = 36$; $1 + 36 = 37$ <- this is not the factor

$2 \times 18 = 36$; $2 + 18 = 20$ <- this is not the factor

$3 \times 12 = 36$; $3 + 12 = 15$ <- this is not the factor

$4 \times 9 = 36$; $4 + 9 = 13$ <- this is not the factor

$6 \times 6 = 36$; $6 + 6 = 12$ <- this IS the factor

$\left(x + 6\right) \left(x + 6\right)$

Jul 15, 2018

${\left(x + 6\right)}^{2}$

#### Explanation:

$\text{ consider the "color(blue)"perfect square } {\left(x \pm a\right)}^{2}$

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

$\text{compare to } {x}^{2} + 12 x + 36$

$\text{with } {a}^{2} = 36 \Rightarrow a = 6$

$\text{and } 2 a x = 2 \times 6 x = 12 x$

$\Rightarrow {x}^{2} + 12 x + 36 = {\left(x + 6\right)}^{2}$

Jul 15, 2018

${\left(x + 6\right)}^{2}$

#### Explanation:

What two numbers sum up to the middle term ($6$), and have a product of the last term ($36$)?

After some trial and error, we arrive at $6$ and $6$, which means we can factor this as

$\left(x + 6\right) \left(x + 6\right)$ or ${\left(x + 6\right)}^{2}$

Another way we could have approached this is that we see that the leading term and the last term are both perfect squares.

With this in mind, we could rewrite the expression as

${\left(x\right)}^{2} + 12 x + {\left(6\right)}^{2}$

Now, it becomes a little more clearer that we can factor this as

${\left(x + 6\right)}^{2}$

Hope this helps!