factorise: x^2 + 6 + 9?

x² + 6x + 9 = (x + 3)(x + 3) = (x + 3)²

(x+3)(x+3)
the square rule is that for the x squared, u find the square root and use this as the start of the algebraic expressions for both sides.
then, the 2nd pronumeral has to have 2 numbers whose sum is the product of the 3rd pronumeral, in this case, 2 numbers adding up to 6 when multiplied should equal 9,
3+3=6 and 3 times 3=9,
this is the perfect square rule for the factorisation of trinomial expressions.
i hoped this helped =)

you basically could use path and blunder, after slightly practice you'll commence seeing them immediately. (x+10)(x+2) (p-4)(p-5) (ok-4)(ok-9) (x+12)(x-a million) (c-10)(c+3) (2x+a million)(2x+a million) (3-D+2)(d+3) (2x+5)(x-2) (5b+3)(b-2) (2c-a million)(c+3) 2t(2t-a million)(-2t+a million)

Question Number 1 :
For this equation x^2 + 6*x + 9 = 0 , answer the following questions :
A. Use factorization to find the root of the equation !

Answer Number 1 :
The equation x^2 + 6*x + 9 = 0 is already in a*x^2+b*x+c=0 form.
As the value is already arranged in a*x^2+b*x+c=0 form, we get the value of a = 1, b = 6, c = 9.

1A. Use factorization to find the root of the equation !
x^2 + 6*x + 9 = 0
<=> ( x + 3 ) * ( x + 3 ) = 0
We get following answers x1 = -3 and x2 = -3

Shouldn`t it be x^2 + 6x + 9 ? ? ?

x^2 + 6x + 9

= (x + 3)^2

x^2 + 6 + 9
=(x+3)^2

x^2 + 6 + 9
= x^2 + 15

I cant do it if its x^2 + 6 + 9
but if its x^2 + 6x + 9
then
it's (x+3) (x+3)
so x + 3 = o so x = 3