To prove that [tex]\(a \cdot b = c\)[/tex] for the polynomial [tex]\(f(x) = x^3 - a x^2 + b x - c\)[/tex], given that the sum of two zeroes (roots) is zero, let's follow these steps:
1. Represent the roots:
- Let the roots of the polynomial be [tex]\(r_1\)[/tex], [tex]\(r_2\)[/tex], and [tex]\(r_3\)[/tex].
- Given that the sum of two zeroes is zero, we can represent the roots as [tex]\(r_1\)[/tex], [tex]\(r\)[/tex], and [tex]\(-r\)[/tex].
2. Use Vieta's formulas:
Vieta's formulas relate the coefficients of the polynomial to sums and products of its roots. For a cubic polynomial [tex]\(x^3 - a x^2 + b x - c\)[/tex]:
- The sum of the roots [tex]\(r_1 + r + (-r)\)[/tex] is equal to the coefficient of [tex]\(x^2\)[/tex] divided by the leading coefficient (with a negative sign, but since our leading coefficient is 1, it remains as the sum):
[tex]\[
r_1 + r + (-r) = r_1 = a
\][/tex]
So, [tex]\(r_1 = a\)[/tex].
- The product of the roots taken one at a time (the constant term [tex]\(c\)[/tex] with a sign change and divided by the leading coefficient):
[tex]\[
r_1 \cdot r \cdot (-r) = -r^2 \cdot r_1 = -c
\][/tex]
Thus,
[tex]\[
-r^2 \cdot a = -c \implies r^2 \cdot a = c
\][/tex]
- The sum of the products of the roots taken two at a time is equal to the coefficient of [tex]\(x\)[/tex] divided by the leading coefficient:
[tex]\[
r_1 \cdot r + r_1 \cdot (-r) + r \cdot (-r) = b
\][/tex]
Simplifying this, we get:
[tex]\[
a \cdot r + a \cdot (-r) + r \cdot (-r) = b
\implies 0 - r^2 = b
\implies -r^2 = b
\][/tex]
So, [tex]\(r^2 = -b\)[/tex].
3. Combine the equations:
We now have two key equations from the above steps:
[tex]\[
r^2 \cdot a = c \quad \text{and} \quad r^2 = -b
\][/tex]
Substitute [tex]\(r^2 = -b\)[/tex] into [tex]\(r^2 \cdot a = c\)[/tex]:
[tex]\[
(-b) \cdot a = c
\][/tex]
Thus, we have:
[tex]\[
a \cdot b = c
\][/tex]
Hence, we have proven that [tex]\(a \cdot b = c\)[/tex].
