Let's solve this problem step-by-step.
1. Identify the given polynomial and find its zeroes:
Given polynomial is [tex]\(2x^2 - 5x - 3\)[/tex]. To find its zeroes, we solve the quadratic equation:
[tex]\[
2x^2 - 5x - 3 = 0
\][/tex]
The zeroes of this polynomial are [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex], which are the roots of the equation.
2. Find the zeroes of [tex]\(2x^2 - 5x - 3\)[/tex]:
We have the zeroes:
[tex]\[
\alpha = -\frac{1}{2} \quad \text{and} \quad \beta = 3
\][/tex]
3. Determine the zeroes of the new polynomial:
It is given that the zeroes of the polynomial [tex]\(x^2 + px + q\)[/tex] are double the zeroes of [tex]\(2x^2 - 5x - 3\)[/tex]. Therefore, if the zeroes of [tex]\(2x^2 - 5x - 3\)[/tex] are [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex], then the zeroes of [tex]\(x^2 + px + q\)[/tex] are [tex]\(2\alpha\)[/tex] and [tex]\(2\beta\)[/tex].
That means:
[tex]\[
2\alpha = 2 \times \left( -\frac{1}{2} \right) = -1 \quad \text{and} \quad 2\beta = 2 \times 3 = 6
\][/tex]
4. Find the sum and product of the zeroes for the new polynomial:
Let the zeroes of [tex]\(x^2 + px + q\)[/tex] be [tex]\(2\alpha\)[/tex] and [tex]\(2\beta\)[/tex]:
[tex]\[
\sum \text{of zeroes of } (x^2 + px + q) = 2\alpha + 2\beta = -1 + 6 = 5
\][/tex]
[tex]\[
\text{product of zeroes of } (x^2 + px + q) = 2\alpha \times 2\beta = (-1) \times 6 = -6
\][/tex]
5. Relate the zeroes to coefficients of the polynomial [tex]\(x^2 + px + q\)[/tex]:
- For a polynomial [tex]\(x^2 + px + q\)[/tex], the sum of the zeroes equals [tex]\(-p\)[/tex].
Therefore:
[tex]\[
-p = 5 \quad \Rightarrow \quad p = -5
\][/tex]
- The product of the zeroes equals [tex]\(q\)[/tex].
Therefore:
[tex]\[
q = -6
\][/tex]
Thus, the values of [tex]\(p\)[/tex] and [tex]\(q\)[/tex] for the polynomial [tex]\(x^2 + px + q\)[/tex] are:
[tex]\[
\boxed{p = -5 \quad \text{and} \quad q = -6}
\][/tex]
