Problem 129 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(P(x)\) and \(Q(x)\) are two polynomials such that \(f(x)=\) \(P\left(x^{3}\right)+x Q\left(x^{3}\right)\) is divisible by \(x^{2}+x+1\), then (A) \(P(x)\) is divisible by \((x-1)\) but \(Q(x)\) is not divisible by \(x-1\) (B) \(Q(x)\) is divisible by \((x-1)\) but \(P(x)\) is not divisible by \(x-1\) (C) Both \(P(x)\) and \(Q(x)\) are divisible by \(x-1\) (D) \(f(x)\) is divisible by \(x-1\)

Step-by-Step Solution

Answer

(B) \(Q(x)\) is divisible by \(x-1\) but \(P(x)\) is not divisible by \(x-1\).

Step 1: Understanding the Problem

We're given a polynomial function \( f(x) = P(x^3) + x Q(x^3) \) which is divisible by \( x^2 + x + 1 \). We need to explore the properties of the polynomials \( P(x) \) and \( Q(x) \), specifically their divisibility by \( x-1 \).

Step 2: Utilizing Polynomial Division Theorem

According to the Polynomial Division Theorem, if a polynomial \( f(x) \) is divisible by another polynomial \( g(x) \), then the roots of \( g(x) \) must also be roots of \( f(x) \). Here, the polynomial \( x^2 + x + 1 \) has roots \( \omega \) and \( \omega^2 \), where \( \omega \) is a primitive cube root of unity.

Step 3: Substituting Roots into the Equation

Substitute \( \omega \) into the equation. We get \( f(\omega) = P(\omega^3) + \omega Q(\omega^3) = 0 \). Since \( \omega^3 = 1 \), this simplifies to \( P(1) + \omega Q(1) = 0 \). Then substitute \( \omega^2 \), getting \( P(1) + \omega^2 Q(1) = 0 \).

Step 4: Solving the System of Equations

From the equations \( P(1) + \omega Q(1) = 0 \) and \( P(1) + \omega^2 Q(1) = 0 \), subtract one from the other to eliminate \( P(1) \), yielding \( \omega Q(1) - \omega^2 Q(1) = 0 \). Simplifying, we find \((\omega - \omega^2)Q(1) = 0\). As \( \omega eq \omega^2 \), it follows that \( Q(1) = 0 \).

Step 5: Implications for Divisibility

Since \( Q(1) = 0 \), \( Q(x) \) is divisible by \( x - 1 \). However, \( P(1) \) is not necessarily zero, implying that \( P(x) \) is not divisible by \( x - 1 \).

Step 6: Drawing Conclusion

Based on the divisibility properties found, the correct option is (B): \( Q(x) \) is divisible by \( x-1 \) but \( P(x) \) is not.

Key Concepts

Roots of Polynomial

Roots of a polynomial are the values of the variable that make the polynomial equal zero. In the context of the given problem, we explore the roots of the polynomial \( x^2 + x + 1 \). These roots are fundamental in determining the divisibility properties of our original polynomial function \( f(x) = P(x^3) + x Q(x^3) \).

For \( x^2 + x + 1 \), the roots are complex numbers, specifically \( \omega \) and \( \omega^2 \), where \( \omega \) is known as a primitive cube root of unity. Knowing the roots helps us understand which values will satisfy the polynomial equation and thus assists in identifying the factors of the polynomial.

In our polynomial \( f(x) \), understanding that these roots \( \omega \) and \( \omega^2 \) must be zeros is quintessential. We substitute these roots back into \( f(x) \) to explore the relationship between \( P(x) \) and \( Q(x) \), unraveling insights into their divisibility properties.

Primitive Cube Root of Unity

The primitive cube root of unity, often denoted as \( \omega \), is a complex number that is a third root of the number one, other than 1 itself. Mathematically, it satisfies \( \omega^3 = 1 \) and \( \omega eq 1 \).

It plays a pivotal role in the polynomial \( x^2 + x + 1 = 0 \) as its roots are \( \omega \) and \( \omega^2 \). These are special numbers because they represent the positions on the complex plane at 120 degrees (\( \omega \)) and 240 degrees (\( \omega^2 \)) counterclockwise from the positive real axis. These are expressed as:

• \( \omega = \frac{-1 + \sqrt{-3}}{2} \)
• \( \omega^2 = \frac{-1 - \sqrt{-3}}{2} \)

The role of \( \omega \) extends to simplifying algebraic manipulation within polynomial equations, allowing us to express and solve complex polynomial divisions efficiently. In our problem, substituting \( \omega \) into the polynomial allows us to derive conditions that help determine divisibility properties of \( P(x) \) and \( Q(x) \).

Polynomial Division Theorem

The Polynomial Division Theorem is a fundamental concept that states if a polynomial \( f(x) \) is divisible by another polynomial \( g(x) \), then any root of \( g(x) \) must be a root of \( f(x) \) as well.

In the given problem, \( f(x) = P(x^3) + xQ(x^3) \) is divisible by \( x^2 + x + 1 \), with roots \( \omega \) and \( \omega^2 \). The theorem tells us that substituting these roots into \( f(x) \) should result in zero. This gives us two equations \( P(1) + \omega Q(1) = 0 \) and \( P(1) + \omega^2 Q(1) = 0 \).

• Setting \( \omega Q(1) - \omega^2 Q(1) = 0 \) allows us to conclude \( Q(1) = 0 \), signifying that \( Q(x) \) is divisible by \( x - 1 \).
• The simplicity of the subtraction involves using the fact that \( \omega^2 = \omega + 1 \), a property emerging from \( \omega^3 = 1 \).

However, for \( P(x) \), we do not derive \( P(1) = 0 \), so \( P(x) \) may not necessarily be divisible by \( x - 1 \). This theorem efficiently guides us through the logical reasoning to determine the divisibility of polynomials.