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For a cubic polynomial t(x) = px^(3) + q...

For a cubic polynomial `t(x) = px^(3) + qx^(2) + rx +s`, which of the following is always true:

A

`s ne 0`

B

`r ne 0`

C

`q ne 0`

D

`p ne 0`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the cubic polynomial \( t(x) = px^3 + qx^2 + rx + s \), we need to determine which statement is always true about this polynomial. ### Step-by-Step Solution: 1. **Identify the Form of the Cubic Polynomial**: The given polynomial is in the standard form of a cubic polynomial, which is \( t(x) = px^3 + qx^2 + rx + s \). 2. **Understand the Leading Coefficient**: In a cubic polynomial, the term with the highest degree (which is \( x^3 \)) is called the leading term. The coefficient of this term is referred to as the leading coefficient. In this case, the leading coefficient is \( p \). 3. **Condition for a Cubic Polynomial**: For a polynomial to be classified as cubic, the leading coefficient must be non-zero. This means \( p \) must not equal zero. If \( p = 0 \), the polynomial would not be cubic; it would instead be a quadratic or lower degree polynomial. 4. **Conclusion**: Therefore, the statement that is always true for the cubic polynomial \( t(x) = px^3 + qx^2 + rx + s \) is that the leading coefficient \( p \) must be non-zero. Hence, \( p \neq 0 \). ### Final Answer: The correct answer is that the leading coefficient \( p \) of the cubic polynomial is always non-zero, i.e., \( p \neq 0 \). ---
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