To find the values of [tex]\( k \)[/tex] and [tex]\( m \)[/tex] when 2 and 3 are zeroes of the polynomial [tex]\( 3x^2 - 2kx + 2m \)[/tex], we can use properties of polynomials and their roots. Specifically, if a polynomial [tex]\( ax^2 + bx + c \)[/tex] has roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex], then the following relationships hold:
1. The sum of the roots: [tex]\(\alpha + \beta = -\frac{b}{a}\)[/tex]
2. The product of the roots: [tex]\(\alpha \beta = \frac{c}{a}\)[/tex]
Given:
- [tex]\(\alpha = 2\)[/tex]
- [tex]\(\beta = 3\)[/tex]
For the polynomial [tex]\( 3x^2 - 2kx + 2m \)[/tex]:
- [tex]\(a = 3\)[/tex]
- [tex]\(b = -2k\)[/tex]
- [tex]\(c = 2m\)[/tex]
### Step 1: Sum of the roots
Using the sum of the roots:
[tex]\[
\alpha + \beta = 2 + 3 = 5
\][/tex]
[tex]\[
5 = -\frac{b}{a}
\][/tex]
Substituting [tex]\( b = -2k \)[/tex] and [tex]\( a = 3 \)[/tex]:
[tex]\[
5 = -\frac{-2k}{3}
\][/tex]
[tex]\[
5 = \frac{2k}{3}
\][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[
5 \cdot 3 = 2k
\][/tex]
[tex]\[
15 = 2k
\][/tex]
[tex]\[
k = \frac{15}{2} = 7.5
\][/tex]
### Step 2: Product of the roots
Using the product of the roots:
[tex]\[
\alpha \beta = 2 \cdot 3 = 6
\][/tex]
[tex]\[
6 = \frac{c}{a}
\][/tex]
Substituting [tex]\( c = 2m \)[/tex] and [tex]\( a = 3 \)[/tex]:
[tex]\[
6 = \frac{2m}{3}
\][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[
6 \cdot 3 = 2m
\][/tex]
[tex]\[
18 = 2m
\][/tex]
[tex]\[
m = \frac{18}{2} = 9
\][/tex]
Therefore, the values are:
[tex]\[
k = 7.5 \quad \text{and} \quad m = 9
\][/tex]
### Summary
When 2 and 3 are zeroes of the polynomial [tex]\( 3x^2 - 2kx + 2m \)[/tex], the values of the constants are:
[tex]\[
k = 7.5
\][/tex]
[tex]\[
m = 9
\][/tex]
