To factor the trinomial [tex]\(y^2 + 4y - 32\)[/tex], we want to express it in the form [tex]\((y + a)(y + b)\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are constants that we need to determine.
Here is the detailed, step-by-step solution:
1. Identify the coefficients: Given the trinomial [tex]\(y^2 + 4y - 32\)[/tex], the coefficient of [tex]\(y^2\)[/tex] (which we will call [tex]\(A\)[/tex]) is 1, the coefficient of [tex]\(y\)[/tex] (which we will call [tex]\(B\)[/tex]) is 4, and the constant term (which we will call [tex]\(C\)[/tex]) is -32.
2. Determine [tex]\(a\)[/tex] and [tex]\(b\)[/tex]: We are looking for two numbers whose product is [tex]\(A \cdot C = 1 \cdot (-32) = -32\)[/tex] and whose sum is [tex]\(B = 4\)[/tex].
Let's list the pairs of factors of [tex]\(-32\)[/tex] and check their sums:
- [tex]\(1 \cdot (-32) = -32\)[/tex] and [tex]\(1 + (-32) = -31\)[/tex]
- [tex]\(-1 \cdot 32 = -32\)[/tex] and [tex]\(-1 + 32 = 31\)[/tex]
- [tex]\(2 \cdot (-16) = -32\)[/tex] and [tex]\(2 + (-16) = -14\)[/tex]
- [tex]\(-2 \cdot 16 = -32\)[/tex] and [tex]\(-2 + 16 = 14\)[/tex]
- [tex]\(4 \cdot (-8) = -32\)[/tex] and [tex]\(4 + (-8) = -4\)[/tex]
- [tex]\(-4 \cdot 8 = -32\)[/tex] and [tex]\(-4 + 8 = 4\)[/tex] (This is our pair!)
So, the numbers we are looking for are [tex]\(-4\)[/tex] and [tex]\(8\)[/tex].
3. Write the factorized form: Now that we have [tex]\(a = -4\)[/tex] and [tex]\(b = 8\)[/tex], we can express the trinomial in its factored form as:
[tex]\[
(y - 4)(y + 8)
\][/tex]
4. Verify the factorization: (Optional)
To ensure our factorization is correct, we can expand [tex]\((y - 4)(y + 8)\)[/tex] to see if it matches the original trinomial:
[tex]\[
(y - 4)(y + 8) = y^2 + 8y - 4y - 32 = y^2 + 4y - 32
\][/tex]
The expanded form matches the original trinomial, verifying that our factorization is correct.
Therefore, the factored form of the trinomial [tex]\(y^2 + 4y - 32\)[/tex] is:
[tex]\[
(y - 4)(y + 8)
\][/tex]
