Understanding Polynomial Divisibility

The question states that the polynomial \(P(x) = x^6 + px^5 + qx^4 - x^2 - x - 3\) is divisible by the polynomial \(D(x) = x^4 - 1\). This means that when \(P(x)\) is divided by \(D(x)\), the remainder is zero.

We can use the property of polynomial divisibility: if a polynomial \(P(x)\) is divisible by another polynomial \(D(x)\), then any root of \(D(x)\) is also a root of \(P(x)\) if the remainder is zero. Alternatively, we can use polynomial long division or compare coefficients after expressing \(P(x)\) as \(D(x) \cdot Q(x)\) for some polynomial \(Q(x)\).

Finding the Roots of the Divisor \(x^4 - 1\)

The divisor is \(D(x) = x^4 - 1\). To find its roots, we set \(x^4 - 1 = 0\), which gives \(x^4 = 1\). The roots of this equation are \(x = 1, x = -1, x = i, x = -i\).

Since \(P(x)\) is divisible by \(x^4 - 1\), the remainder must be 0 when \(P(x)\) is divided by \(x^4 - 1\). This implies that \(P(x)\) must be equal to \((x^4 - 1) \cdot Q(x)\) for some polynomial \(Q(x)\).

Setting \(x^6 + px^5 + qx^4 - x^2 - x - 3 = (x^4 - 1)Q(x)\).

Using the Remainder Theorem with Roots

A more general form of the Remainder Theorem states that if a polynomial \(P(x)\) is divided by \((x-a)\), the remainder is \(P(a)\). If \(P(x)\) is divisible by \((x-a)\), then \(P(a)=0\). If \(P(x)\) is divisible by \((x^4 - 1)\), it must be divisible by each of its factors, and the remainder when dividing by \(x^4 - 1\) is zero.

Let's evaluate \(P(x)\) at some of the roots of \(x^4 - 1\).

• For \(x=1\):

  \(P(1) = 1^6 + p(1)^5 + q(1)^4 - 1^2 - 1 - 3\) \(P(1) = 1 + p + q - 1 - 1 - 3\) \(P(1) = p + q - 4\) Since \(P(x)\) is divisible by \(x^4 - 1\), the remainder is 0. So, \(P(1) = 0\). \(p + q - 4 = 0 \implies p + q = 4 \quad (Equation\; 1)\)

• For \(x=-1\):

  \(P(-1) = (-1)^6 + p(-1)^5 + q(-1)^4 - (-1)^2 - (-1) - 3\) \(P(-1) = 1 + p(-1) + q(1) - (1) - (-1) - 3\) \(P(-1) = 1 - p + q - 1 + 1 - 3\) \(P(-1) = -p + q - 2\) Since \(P(x)\) is divisible by \(x^4 - 1\), the remainder is 0. So, \(P(-1) = 0\). \(-p + q - 2 = 0 \implies -p + q = 2 \quad (Equation\; 2)\)

We now have a system of two linear equations involving \(p\) and \(q\):

Solving for \(p\) and \(q\)

We can solve this system of equations by adding the two equations:

Now substitute the value of \(q\) into Equation 1:

So, we found that \(p=1\) and \(q=3\).

Calculating \(p^2 + q^2\)

The question asks for the value of \(p^2 + q^2\).

Alternative Method: Comparing Coefficients using Polynomial Long Division

If \(x^6 + px^5 + qx^4 - x^2 - x - 3\) is divisible by \(x^4 - 1\), the polynomial can be written as \((x^4 - 1) \cdot Q(x)\). By looking at the degrees, \(Q(x)\) must be a quadratic polynomial, say \(Q(x) = ax^2 + bx + c\).

Comparing the coefficients of the powers of \(x\) on both sides:

• Coefficient of \(x^6\): \(1 = a \implies a = 1\)
• Coefficient of \(x^5\): \(p = b \implies b = p\)
• Coefficient of \(x^4\): \(q = c \implies c = q\)
• Coefficient of \(x^3\): \(0 = 0\) (There is no \(x^3\) term on either side from \(ax^6 + bx^5 + cx^4 - ax^2 - bx - c\))
• Coefficient of \(x^2\): \(-1 = -a\)
• Coefficient of \(x^1\): \(-1 = -b\)
• Constant term: \(-3 = -c\)

From the coefficients of \(x^2, x^1\), and the constant term:

• \(-1 = -a \implies a = 1\) (This matches the coefficient of \(x^6\))
• \(-1 = -b \implies b = 1\)
• \(-3 = -c \implies c = 3\)

Comparing these values with the coefficients of \(x^5\) and \(x^4\):

• From \(p = b\), we get \(p = 1\).
• From \(q = c\), we get \(q = 3\).

This method also gives \(p=1\) and \(q=3\).

Now calculate \(p^2 + q^2\):

Both methods confirm that the value of \(p^2 + q^2\) is 10.

Revision Table: Key Concepts in Polynomial Divisibility

Additional Information on Polynomials

Polynomials are fundamental algebraic expressions used widely in mathematics and science. Understanding how to perform operations like division, and concepts like roots and factors, is crucial for solving various problems.

• Degree of a Polynomial: The highest power of the variable in the polynomial. In the question, \(x^6 + px^5 + qx^4 - x^2 - x - 3\) is of degree 6, and \(x^4 - 1\) is of degree 4.
• Factor Theorem: A polynomial \(P(x)\) has a factor \((x-a)\) if and only if \(P(a)=0\). This is a direct consequence of the Remainder Theorem.
• Complex Roots: Polynomials can have complex roots (involving \(i\), where \(i^2 = -1\)). In this problem, \(x^4 - 1\) has two real roots (\(1, -1\)) and two complex roots (\(i, -i\)). Using real roots was sufficient to find \(p\) and \(q\).

Polynomial long division is a systematic algorithm for dividing polynomials, similar to numerical long division. It can always be used to find the quotient and remainder when one polynomial is divided by another. In cases of exact divisibility, the remainder is zero.