Polynomial Divisibility Solution

We are given that the polynomial \(P(x) = x^4 + px^3 + qx^2 + x + 6\) is divisible by \(D(x) = x^2 - x - 6\). We need to find the value of \(p\). Divisibility implies that the roots of the divisor polynomial \(D(x)\) are also roots of the polynomial \(P(x)\).

1. Find the Roots of the Divisor

First, find the roots of the divisor \(D(x) = x^2 - x - 6\). Set \(D(x) = 0\) and solve for \(x\):

$x^2 - x - 6 = 0$

Factorizing the quadratic equation:

$\implies (x - 3)(x + 2) = 0$

The roots are \(x = 3\) and \(x = -2\).

2. Apply the Divisibility Condition

Since \(P(x)\) is divisible by \(D(x)\), the roots of \(D(x)\) must also be roots of \(P(x)\). Therefore, \(P(3) = 0\) and \(P(-2) = 0\).

3. Substitute Roots into P(x)

Substitute \(x = 3\) into \(P(x)\):

$P(3) = (3)^4 + p(3)^3 + q(3)^2 + (3) + 6 = 0$

$\implies 81 + 27p + 9q + 3 + 6 = 0$

$\implies 27p + 9q + 90 = 0$

Divide by 9:

$\implies 3p + q + 10 = 0 \quad (1)$

Substitute \(x = -2\) into \(P(x)\):

$P(-2) = (-2)^4 + p(-2)^3 + q(-2)^2 + (-2) + 6 = 0$

$\implies 16 + p(-8) + q(4) - 2 + 6 = 0$

$\implies 16 - 8p + 4q + 4 = 0$

$\implies -8p + 4q + 20 = 0$

Divide by 4:

$\implies -2p + q + 5 = 0 \quad (2)$

4. Solve the System of Equations

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

1. $3p + q + 10 = 0$
2. $-2p + q + 5 = 0$

Subtract equation (2) from equation (1) to eliminate \(q\):

$(3p + q + 10) - (-2p + q + 5) = 0 - 0$

$\implies 3p + q + 10 + 2p - q - 5 = 0$

$\implies 5p + 5 = 0$

$\implies 5p = -5$

$\implies p = -1$

5. Conclusion

The value of \(p\) is \(-1\).