To solve the problem, we first need to understand what it means for 1 to be a zero of the polynomials [tex]\(ay^2 + ay + 3\)[/tex] and [tex]\(y + a + b\)[/tex].
### Step-by-Step Solution:
1. First Polynomial: [tex]\(ay^2 + ay + 3\)[/tex]
- Since 1 is a zero of the polynomial, substituting [tex]\(y = 1\)[/tex] should make the polynomial equal to zero.
[tex]\[
a(1)^2 + a(1) + 3 = 0
\][/tex]
Simplifying, we get:
[tex]\[
a + a + 3 = 0
\][/tex]
Combining like terms, we have:
[tex]\[
2a + 3 = 0
\][/tex]
Solving for [tex]\(a\)[/tex], we subtract 3 from both sides:
[tex]\[
2a = -3
\][/tex]
Then, divide by 2:
[tex]\[
a = -\frac{3}{2}
\][/tex]
2. Second Polynomial: [tex]\(y + a + b\)[/tex]
- Again, since 1 is a zero of this polynomial, substituting [tex]\(y = 1\)[/tex] should make the polynomial equal to zero.
[tex]\[
1 + a + b = 0
\][/tex]
We already determined that [tex]\(a = -\frac{3}{2}\)[/tex]. Substituting this value into the equation:
[tex]\[
1 + \left(-\frac{3}{2}\right) + b = 0
\][/tex]
Simplifying inside the parentheses:
[tex]\[
1 - \frac{3}{2} + b = 0
\][/tex]
Converting 1 to a fraction:
[tex]\[
\frac{2}{2} - \frac{3}{2} + b = 0
\][/tex]
Combining the fractions:
[tex]\[
\frac{2 - 3}{2} + b = 0
\][/tex]
Simplifying the fractions:
[tex]\[
-\frac{1}{2} + b = 0
\][/tex]
Solving for [tex]\(b\)[/tex], we add [tex]\( \frac{1}{2} \)[/tex] to both sides:
[tex]\[
b = \frac{1}{2}
\][/tex]
3. Finding the Product [tex]\(ab\)[/tex]
- We now have [tex]\(a = -\frac{3}{2}\)[/tex] and [tex]\(b = \frac{1}{2}\)[/tex]. To find [tex]\(ab\)[/tex]:
[tex]\[
ab = \left( -\frac{3}{2} \right) \left( \frac{1}{2} \right)
\][/tex]
Multiplying the fractions:
[tex]\[
ab = \frac{-3 \cdot 1}{2 \cdot 2}
\][/tex]
Simplifying the numerator and the denominator:
[tex]\[
ab = \frac{-3}{4}
\][/tex]
Thus, the value of [tex]\(ab\)[/tex] is:
[tex]\[
ab = -0.75
\][/tex]
So, the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( ab \)[/tex] are [tex]\(-1.5\)[/tex], [tex]\(0.5\)[/tex], and [tex]\(-0.75\)[/tex], respectively.
