Let Alpha and Beta be the zeros of
the cubic polynomial x^3 + ax^2 +
bx + c satisfying the relation
Alpha * Beta = 0. Prove that : c^2 +
ac + b + 1 = 0.
Please work the complete solution.
To prove the given relation, we need to make use of the Vieta's formulas for the zeros of a polynomial.
Vieta's formulas state that for a cubic polynomial of the form ax^3 + bx^2 + cx + d, the sum of the zeros (Alpha + Beta + Gamma) is equal to -b/a, the sum of the product of every possible pair of zeros (Alpha * Beta + Beta * Gamma + Gamma * Alpha) is equal to c/a, and the product of the zeros (Alpha * Beta * Gamma) is equal to -d/a.
In this case, we are given that Alpha * Beta = 0. Using Vieta's formulas, we can determine the relation between the coefficients of the polynomial.
According to Vieta's formulas, we have:
Alpha + Beta + Gamma = -a (Equation 1)
Alpha * Beta + Beta * Gamma + Gamma * Alpha = b (Equation 2)
Alpha * Beta * Gamma = -c (Equation 3)
Since Alpha * Beta = 0, one of the zeros must be 0. Without loss of generality, let's assume Alpha = 0. This means Beta * Gamma = -c according to Equation 3.
Substituting Alpha = 0 and Alpha * Beta = 0 into Equation 1 and Equation 2, we get:
Beta + Gamma = -a (Equation 4)
Beta * Gamma = b (Equation 5)
Now, we can express Gamma in terms of Beta using Equations 4 and 5:
Gamma = -(a + Beta) (Equation 6)
Substituting this value of Gamma into Equation 5, we can simplify the equation:
Beta * Gamma = b
Beta * (-(a + Beta)) = b
-aBeta - Beta^2 = b
Beta^2 + aBeta + b = 0 (Equation 7)
Finally, we are given that Gamma is the remaining zero of the polynomial. Therefore, the polynomial can be written as:
x^3 + ax^2 + bx + c = x(x - Beta)(x - Gamma)
= x(x^2 - (Beta + Gamma)x + Beta * Gamma)
= x(x^2 - (Beta + Gamma)x + b) (using Equation 5)
Comparing this to the original cubic polynomial, we can see that:
c = b, and a = -(Beta + Gamma)
Substituting these values into the equation to prove, we get:
c^2 + ac + b + 1 = b^2 + a(b - a) + b + 1
= b^2 + ab - a^2 + b + 1
= b^2 + ab + b + 1 - (a^2 + a)
= b(b + a) + 1 - a(a + 1)
= (Beta^2 + aBeta) + 1 - (a^2 + a) (using Equation 7)
= 0 + 1 - 0 (from Equation 7)
= 1
Thus, we have c^2 + ac + b + 1 = 0, thereby proving the given relation.