$x^4 + px^3 + qx^2 + x + 6$ is divisible by $x^2 - x - 6$
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 \(x\):
Factorizing the quadratic equation:
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 \(P(x)\):
Divide by 9:
Substitute \(x = -2\) into \(P(x)\):
Divide by 4:
4. Solve the System of Equations
We now have a system of two linear equations with two variables, \(p\) and \(q\):
Subtract equation (2) from equation (1) to eliminate \(q\):
5. Conclusion
The value of \(p\) is \(-1\).
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