Sure, let's walk through the process of factoring the given trinomial and then verify the factorization using FOIL multiplication.
We are given the trinomial:
[tex]\[ y^2 - 12y + 35 \][/tex]
To factor this trinomial, we need to find two numbers that multiply to the constant term (35) and add up to the coefficient of the middle term (-12).
The two numbers that fit these criteria are -7 and -5, because:
[tex]\[ (-7) \times (-5) = 35 \][/tex]
[tex]\[ (-7) + (-5) = -12 \][/tex]
Thus, we can express the trinomial as a product of two binomials:
[tex]\[ y^2 - 12y + 35 = (y - 7)(y - 5) \][/tex]
To verify this factorization, we'll use the FOIL (First, Outer, Inner, Last) method to multiply the binomials and ensure we get back the original trinomial.
FOIL Method:
1. First: Multiply the first terms in each binomial:
[tex]\[ y \cdot y = y^2 \][/tex]
2. Outer: Multiply the outer terms in the binomials:
[tex]\[ y \cdot (-5) = -5y \][/tex]
3. Inner: Multiply the inner terms in the binomials:
[tex]\[ (-7) \cdot y = -7y \][/tex]
4. Last: Multiply the last terms in each binomial:
[tex]\[ (-7) \cdot (-5) = 35 \][/tex]
Now, combine all the products:
[tex]\[ y^2 + (-5y) + (-7y) + 35 \][/tex]
[tex]\[ y^2 - 5y - 7y + 35 \][/tex]
[tex]\[ y^2 - 12y + 35 \][/tex]
Since we have successfully returned to the original trinomial, our factorization is correct.
Therefore, the correct choice is:
[tex]\[ \boxed{A} \quad y^2 - 12y + 35 = (y - 7)(y - 5) \][/tex]
The trinomial [tex]$y^2 - 12y + 35$[/tex] is not prime, as it can be factored into [tex]$(y - 7)(y - 5)$[/tex].
