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Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes is
- a)-c/a
- b)c/a
- c)0
- d)-b/a
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + ...
Let p(x) =ax3 + bx2 + cx + d
Given that, one of the zeroes of the cubic polynomial p(x) is zero.
Let α, β and γ are the zeroes of cubic polynomial p(x), where a = 0.
We know that,
Given that, one of the zeroes of the cubic polynomial p(x) is zero.
Let α, β and γ are the zeroes of cubic polynomial p(x), where a = 0.
We know that,
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Most Upvoted Answer
Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + ...
Community Answer
Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + ...
To understand why the correct answer is option 'B', let's first review some properties of cubic polynomials and their zeroes.
A cubic polynomial is a polynomial of degree 3, which means it can have at most 3 zeroes. These zeroes can be real or complex numbers. If one of the zeroes of a cubic polynomial is known, we can use this information to find the other two zeroes.
Now, let's denote the known zero as "x = 0". Since it is a zero of the polynomial, we can write the polynomial equation as:
ax^3 + bx^2 + cx + d = 0
We know that x = 0 is a zero, so substituting this value into the equation, we get:
a(0)^3 + b(0)^2 + c(0) + d = 0
0 + 0 + 0 + d = 0
d = 0
Therefore, we have found that d = 0. Now, let's substitute this value back into the polynomial equation:
ax^3 + bx^2 + cx + 0 = 0
ax^3 + bx^2 + cx = 0
We can now factor out x from the equation:
x(ax^2 + bx + c) = 0
Since x = 0 is one of the zeroes, we can divide the equation by x:
ax^2 + bx + c = 0
Now, we have a quadratic equation. The product of the other two zeroes can be found using the fact that the product of the zeroes of a quadratic equation is equal to the constant term divided by the coefficient of the highest power of x. In this case, the constant term is 'c' and the coefficient of the highest power of x is 'a'.
Product of other two zeroes = c/a
Therefore, the product of the other two zeroes is c/a. Hence, the correct answer is option 'B' (c/a).
A cubic polynomial is a polynomial of degree 3, which means it can have at most 3 zeroes. These zeroes can be real or complex numbers. If one of the zeroes of a cubic polynomial is known, we can use this information to find the other two zeroes.
Now, let's denote the known zero as "x = 0". Since it is a zero of the polynomial, we can write the polynomial equation as:
ax^3 + bx^2 + cx + d = 0
We know that x = 0 is a zero, so substituting this value into the equation, we get:
a(0)^3 + b(0)^2 + c(0) + d = 0
0 + 0 + 0 + d = 0
d = 0
Therefore, we have found that d = 0. Now, let's substitute this value back into the polynomial equation:
ax^3 + bx^2 + cx + 0 = 0
ax^3 + bx^2 + cx = 0
We can now factor out x from the equation:
x(ax^2 + bx + c) = 0
Since x = 0 is one of the zeroes, we can divide the equation by x:
ax^2 + bx + c = 0
Now, we have a quadratic equation. The product of the other two zeroes can be found using the fact that the product of the zeroes of a quadratic equation is equal to the constant term divided by the coefficient of the highest power of x. In this case, the constant term is 'c' and the coefficient of the highest power of x is 'a'.
Product of other two zeroes = c/a
Therefore, the product of the other two zeroes is c/a. Hence, the correct answer is option 'B' (c/a).
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Given that one of the zeroes of the cubic polynomial ax3 + bx2 + cx + d is zero, the product of the other two zeroes isa)-c/ab)c/ac)0d)-b/aCorrect answer is option 'B'. Can you explain this answer?
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