Question

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

Answer 1

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`:

`(x )(x )`

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.

`(x + )(x + )`

Now we need to determine the factors which multiply to 6 and also add to 5:

`1 xx 6 = 6`; `1 + 6 = 7` <- this is not the factor

`2 xx 3 = 6`; `2 + 3 = 5` <- this IS the factor

`(x + 2)(x + 3)`

Answer 2

Factorize of polynomial `x^2+5x+6 = (x+2)*(x+3)`

Explanation:

`x^2 + (a+b)x + ab = (x+a)*(x+b)`

`x^2+5x+6 = (x+a)*(x+b)`

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

Answer 3

(x+3)(x+2)

Explanation:

`color(purple)("First Answer:")`

Using prime factorization of `6`

`+6=2xx3=1xx6=-2xx-3=-1xx-6`

since the coefficient of `x` is positive

`color(red)("the answers "-2xx-3 " and " -1xx-6 " are refused ")`

`color(blue)("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` `rarr` `color(red)("refused")`

`2+3=5` `rarr` `color(blue)("accepted")`

so it will be

`x^2+5x+6=(x+2)(x+3)`

`color(purple)("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(x-(-bcolor(red)(+)sqrt(b^2-4ac))/(2a))(x-(-bcolor(red)(-)sqrt(b^2-4ac))/(2a))`

so for your question

`x^2+5x+6`

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

`x=(-5+-sqrt(25-4xx1xx6))/(2xx1)=(-5+-1)/2`

and by using it in the formula

`1*(x-(-5+1)/2)(x-(-5-1)/2)`

`=(x+2)(x+3)`

Answer 4

`(x+2)(x+3)`

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*3=6` (Last term)

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

`(x+2)(x+3)`

Hope this helps!