Factor Theorem: Finding P And Q In A Cubic Polynomial

Let's dive into a fascinating problem involving polynomial functions, the factor theorem, and a bit of algebraic manipulation. We're given a cubic polynomial $f(x) = x^3 + (p+1)x^2 - 18x + q$, where $p$ and $q$ are integers. Our mission is to uncover the values of $p$ and $q$ using the information that $(x-4)$ and $(x+p)$ are factors of $f(x)$, with the added condition that $p > 0$. This exploration will not only enhance your understanding of polynomial factorization but also showcase the power of the factor theorem in solving such problems.

(a) Showing That $16p + q + 8 = 0$

Since $(x-4)$ is a factor of $f(x)$, we can apply the factor theorem, which states that if $(x-a)$ is a factor of $f(x)$, then $f(a) = 0$. In our case, $a = 4$, so we have $f(4) = 0$. Let's substitute $x = 4$ into the expression for $f(x)$:

$f(4) = (4)^3 + (p+1)(4)^2 - 18(4) + q = 0$ $64 + 16(p+1) - 72 + q = 0$ $64 + 16p + 16 - 72 + q = 0$ $16p + q + 8 = 0$

Thus, we've successfully shown that $16p + q + 8 = 0$. This equation provides a crucial relationship between $p$ and $q$, which we'll use later to solve for their values. Understanding this step is paramount as it sets the foundation for the subsequent parts of the problem. The factor theorem is a cornerstone in polynomial algebra, and this application demonstrates its practical utility. Remember, this theorem allows us to connect the roots of a polynomial with its factors, providing a powerful tool for solving algebraic equations. This connection is especially useful when dealing with higher-degree polynomials where direct factorization might be challenging. The equation $16p + q + 8 = 0$ now serves as a constraint that $p$ and $q$ must satisfy, narrowing down the possible solutions.

(b) Showing That $p^2 + p - 18 + q/p = 0$ which implies $p^2 + 18 - p + q = 0$

Now, we're given that $(x+p)$ is also a factor of $f(x)$. Applying the factor theorem again, we have $f(-p) = 0$. Substituting $x = -p$ into the expression for $f(x)$:

$f(-p) = (-p)^3 + (p+1)(-p)^2 - 18(-p) + q = 0$ $-p^3 + (p+1)p^2 + 18p + q = 0$ $-p^3 + p^3 + p^2 + 18p + q = 0$ $p^2 + 18p + q = 0$

So, we've shown that $p^2 + 18p + q = 0$. This equation gives us another relationship between $p$ and $q$, which, combined with the previous equation, will allow us to solve for $p$ and $q$ uniquely. The fact that $(x+p)$ is a factor introduces another constraint on the values of $p$ and $q$, enabling us to create a system of equations. Solving this system will lead us to the specific values of $p$ and $q$ that satisfy both conditions. The importance of recognizing and utilizing the factor theorem cannot be overstated. It's a fundamental concept that simplifies polynomial problems and provides a direct link between factors and roots. By applying the factor theorem twice, we've established two equations that relate $p$ and $q$, paving the way for finding their exact values. This process exemplifies how mathematical theorems can be used as tools to dissect and solve complex problems systematically.

(c) Finding the Values of $p$ and $q$

We have two equations:

1. $16p + q + 8 = 0$
2. $p^2 + 18p + q = 0$

We can solve this system of equations by substitution or elimination. Let's use elimination. Subtract equation (1) from equation (2):

$(p^2 + 18p + q) - (16p + q + 8) = 0$ $p^2 + 18p + q - 16p - q - 8 = 0$ $p^2 + 2p - 8 = 0$

Now, we have a quadratic equation in terms of $p$. Let's factor it:

$(p+4)(p-2) = 0$

This gives us two possible values for $p$: $p = -4$ or $p = 2$. However, we're given that $p > 0$, so we must have $p = 2$.

Now that we know $p = 2$, we can substitute it back into either equation (1) or (2) to solve for $q$. Let's use equation (1):

$16(2) + q + 8 = 0$ $32 + q + 8 = 0$ $q + 40 = 0$ $q = -40$

Therefore, the values of $p$ and $q$ are $p = 2$ and $q = -40$. This concludes our solution. The process of solving for $p$ and $q$ involved a combination of algebraic manipulation and problem-solving techniques. We began by applying the factor theorem to establish two equations relating $p$ and $q$. Then, we used elimination to reduce the system of equations to a single quadratic equation in terms of $p$. By solving the quadratic equation and considering the constraint $p > 0$, we found the value of $p$. Finally, we substituted the value of $p$ back into one of the original equations to find the value of $q$. This systematic approach demonstrates how algebraic problems can be tackled by breaking them down into smaller, manageable steps. The values $p = 2$ and $q = -40$ satisfy both conditions given in the problem, confirming the validity of our solution. In summary, the factor theorem played a crucial role in setting up the equations, while algebraic manipulation and problem-solving skills were essential for finding the values of $p$ and $q$.

In conclusion, we successfully found the values of $p$ and $q$ by applying the factor theorem and solving the resulting system of equations. This problem showcases the beauty and power of algebraic techniques in solving polynomial equations. For further reading on the factor theorem, you can check out this resource on Khan Academy.