You mean from (y - 3) * (y^2 + 3y + 9) + (y - 3) = 0 Well, let's rearrange the (y - 3) * (y^2 + 3y + 9) first into... (y^2 + 3y + 9) (y - 3) + (y - 3) = 0 We can notice from the two adding terms that they both have (y - 3), so we can treat them as like terms. Kind of like 3x + x = 4x. Notice that I treated the 'x' as 1x. This is because 1*x = x. Whenever a term is by itself, it is automatically implied that the term has 1 as its coefficient when adding or subtracting it with like terms like with a + 2a. In the same fashion, (y^2 + 3y + 9) is treated as the coefficient for (y^2 + 3y + 9) (y - 3). Since the second term is (y - 3) and the term is by itself, it can be treated as (1) (y - 3) because 1* (y - 3) = (y - 3). So we have (y^2 + 3y + 9) (y - 3) + (1) (y - 3) = 0 Adding the two terms and you get (y^2 + 3y + 10) (y - 3) = 0
how the f is y-3=1 at min 4?
factorizing.
y-3 = a
y² + 3y + 9 = b
ab + a = 0
a(b + 1) = 0
You mean from (y - 3) * (y^2 + 3y + 9) + (y - 3) = 0
Well, let's rearrange the (y - 3) * (y^2 + 3y + 9) first into...
(y^2 + 3y + 9) (y - 3) + (y - 3) = 0
We can notice from the two adding terms that they both have (y - 3), so we can treat
them as like terms. Kind of like 3x + x = 4x. Notice that I treated the 'x' as 1x. This is because 1*x = x. Whenever a term is by itself, it is automatically implied that the term has 1 as its coefficient when adding or subtracting it with like terms like with a + 2a.
In the same fashion, (y^2 + 3y + 9) is treated as the coefficient for (y^2 + 3y + 9) (y - 3). Since the second term is (y - 3) and the term is by itself, it can be treated as (1) (y - 3) because 1* (y - 3) = (y - 3). So we have
(y^2 + 3y + 9) (y - 3) + (1) (y - 3) = 0
Adding the two terms and you get
(y^2 + 3y + 10) (y - 3) = 0