# How do you factor x^2 + 5x + 6?

Jun 20, 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 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 6 and also add to 5:

$1 \times 6 = 6$; $1 + 6 = 7$ <- this is not the factor

$2 \times 3 = 6$; $2 + 3 = 5$ <- this IS the factor

$\left(x + 2\right) \left(x + 3\right)$

Jun 20, 2018

Factorize of polynomial ${x}^{2} + 5 x + 6 = \left(x + 2\right) \cdot \left(x + 3\right)$

#### Explanation:

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

${x}^{2} + 5 x + 6 = \left(x + a\right) \cdot \left(x + b\right)$

Equation system:
$1. a + b = 5$
$2. a b = 6$
Then $a = 2$ and $b = 3$

Jun 20, 2018

(x+3)(x+2)

#### Explanation:

$\textcolor{p u r p \le}{\text{First Answer:}}$

Using prime factorization of $6$

$+ 6 = 2 \times 3 = 1 \times 6 = - 2 \times - 3 = - 1 \times - 6$

since the coefficient of $x$ is positive

$\textcolor{red}{\text{the answers "-2xx-3 " and " -1xx-6 " are refused }}$

$\textcolor{b l u e}{\text{and the answers "2xx3 " and " 1xx6 " are accepted }}$

and then we try adding the two numbers to get the coefficient of $x$

$1 + 6 = 7$$\rightarrow$$\textcolor{red}{\text{refused}}$

$2 + 3 = 5$$\rightarrow$$\textcolor{b l u e}{\text{accepted}}$

so it will be

${x}^{2} + 5 x + 6 = \left(x + 2\right) \left(x + 3\right)$

$\textcolor{p u r p \le}{\text{The Second Answer}}$

if you have a polynomial function
color(blue)(ax^2+bx+c=0

the solution set will be

color(green)(x=(-bcolor(red)(+-)sqrt(b^2-4ac))/(2a)

and using it like this

$a \left(x - \frac{- b \textcolor{red}{+} \sqrt{{b}^{2} - 4 a c}}{2 a}\right) \left(x - \frac{- b \textcolor{red}{-} \sqrt{{b}^{2} - 4 a c}}{2 a}\right)$

${x}^{2} + 5 x + 6$

$a = 1 , b = 5 , c = 6$

$x = \frac{- 5 \pm \sqrt{25 - 4 \times 1 \times 6}}{2 \times 1} = \frac{- 5 \pm 1}{2}$

and by using it in the formula

$1 \cdot \left(x - \frac{- 5 + 1}{2}\right) \left(x - \frac{- 5 - 1}{2}\right)$

$= \left(x + 2\right) \left(x + 3\right)$

Jun 20, 2018

$\left(x + 2\right) \left(x + 3\right)$

#### Explanation:

The key realization is that we need to think of two numbers that add up to the middle term, and have a product of the middle term.

What two numbers sum up to $5$ and have a product of $6$?

After some trial and error, we arrive at $2$ and $3$, because

$2 + 3 = 5$ (Middle term) and

$2 \cdot 3 = 6$ (Last term)

These will be our factors. So we can write our expression as

$\left(x + 2\right) \left(x + 3\right)$

Hope this helps!