Understanding the Problem: Polynomial Factors

The question states that the expression $(x - 1)^3$ is a factor of the polynomial $P(x) = x^4 + \alpha x^3 + \beta x^2 + \gamma x - 1$. We need to find the other factor of this polynomial.

Since $P(x)$ is a polynomial of degree 4 and $(x - 1)^3$ is a factor of degree 3, the other factor must be a polynomial of degree $4 - 3 = 1$. This means the other factor is a linear expression of the form $(x - r)$ for some constant $r$, or simply $kx+c$. Since the leading coefficient of $P(x)$ is 1 and the leading coefficient of $(x-1)^3$ is 1, the leading coefficient of the other factor must also be 1, so it is of the form $(x-r)$.

If $(x-1)^3$ is a factor, it means that $x=1$ is a root of the polynomial $P(x)$ with a multiplicity of at least 3.

Determining the Polynomial Coefficients ($\alpha, \beta, \gamma$)

For $x=1$ to be a root of multiplicity at least 3, the polynomial $P(x)$ and its first two derivatives must be zero at $x=1$. That is, $P(1) = 0$, $P'(1) = 0$, and $P''(1) = 0$.

Let's find the derivatives of $P(x)$:

• $P(x) = x^4 + \alpha x^3 + \beta x^2 + \gamma x - 1$
• $P'(x) = \frac{d}{dx}(x^4 + \alpha x^3 + \beta x^2 + \gamma x - 1) = 4x^3 + 3\alpha x^2 + 2\beta x + \gamma$
• $P''(x) = \frac{d}{dx}(4x^3 + 3\alpha x^2 + 2\beta x + \gamma) = 12x^2 + 6\alpha x + 2\beta$

Now, we set the polynomial and its derivatives to zero at $x=1$:

• $P(1) = (1)^4 + \alpha(1)^3 + \beta(1)^2 + \gamma(1) - 1 = 1 + \alpha + \beta + \gamma - 1 = \alpha + \beta + \gamma = 0$ (Equation 1)
• $P'(1) = 4(1)^3 + 3\alpha(1)^2 + 2\beta(1) + \gamma = 4 + 3\alpha + 2\beta + \gamma = 0$ (Equation 2)
• $P''(1) = 12(1)^2 + 6\alpha(1) + 2\beta = 12 + 6\alpha + 2\beta = 0$ (Equation 3)

We now have a system of three linear equations with three variables $\alpha, \beta, \gamma$. Let's solve this system.

From Equation 3:

\begin{equation*} 12 + 6\alpha + 2\beta = 0 \implies 6\alpha + 2\beta = -12 \implies 3\alpha + \beta = -6 \text{ (Equation 4)} \end{equation*}

Subtract Equation 1 from Equation 2:

\begin{equation*} (4 + 3\alpha + 2\beta + \gamma) - (\alpha + \beta + \gamma) = 0 - 0 \end{equation*}

\begin{equation*} 4 + 2\alpha + \beta = 0 \implies 2\alpha + \beta = -4 \text{ (Equation 5)} \end{equation*}

Now we solve the system of Equations 4 and 5 for $\alpha$ and $\beta$:

\begin{align*} 3\alpha + \beta &= -6 \\ 2\alpha + \beta &= -4 \end{align*}

Subtract Equation 5 from Equation 4:

\begin{equation*} (3\alpha + \beta) - (2\alpha + \beta) = -6 - (-4) \end{equation*}

\begin{equation*} \alpha = -6 + 4 = -2 \end{equation*}

Substitute the value of $\alpha = -2$ into Equation 5:

\begin{equation*} 2(-2) + \beta = -4 \end{equation*}

\begin{equation*} -4 + \beta = -4 \implies \beta = 0 \end{equation*}

Substitute the values of $\alpha = -2$ and $\beta = 0$ into Equation 1:

\begin{equation*} \alpha + \beta + \gamma = 0 \end{equation*}

\begin{equation*} -2 + 0 + \gamma = 0 \implies \gamma = 2 \end{equation*}

So, the coefficients are $\alpha = -2$, $\beta = 0$, and $\gamma = 2$. The polynomial is $P(x) = x^4 - 2x^3 + 0x^2 + 2x - 1$, which simplifies to $P(x) = x^4 - 2x^3 + 2x - 1$.

Finding the Remaining Factor using Roots

We know that $P(x)$ is a polynomial of degree 4. If $(x-1)^3$ is a factor, it means $x=1$ is a root with multiplicity 3. A polynomial of degree 4 has exactly 4 roots (counting multiplicity). Let the four roots be $r_1, r_2, r_3, r_4$. We know three of the roots are $1, 1, 1$. Let the fourth root be $r_4$.

For a general quartic polynomial $ax^4 + bx^3 + cx^2 + dx + e = 0$, the sum of the roots is given by $-b/a$.

In our polynomial $P(x) = x^4 - 2x^3 + 2x - 1$, the coefficients are $a=1$, $b=-2$, $c=0$, $d=2$, $e=-1$.

The sum of the roots is:

\begin{equation*} r_1 + r_2 + r_3 + r_4 = -\frac{b}{a} \end{equation*}

\begin{equation*} 1 + 1 + 1 + r_4 = -\frac{-2}{1} \end{equation*}

\begin{equation*} 3 + r_4 = 2 \end{equation*}

Solving for $r_4$:

\begin{equation*} r_4 = 2 - 3 = -1 \end{equation*}

The fourth root is $-1$. If $-1$ is a root, then $(x - (-1))$ is a factor. This means $(x+1)$ is the other factor.

Verifying the Factors

The polynomial is $x^4 - 2x^3 + 2x - 1$. The factors are $(x-1)^3$ and $(x+1)$. Let's multiply them to check:

First, expand $(x-1)^3$: $(x-1)^3 = x^3 - 3x^2(1) + 3x(1)^2 - 1^3 = x^3 - 3x^2 + 3x - 1$.

Now, multiply $(x^3 - 3x^2 + 3x - 1)$ by $(x+1)$:

\begin{align*} (x^3 - 3x^2 + 3x - 1)(x+1) &= x(x^3 - 3x^2 + 3x - 1) + 1(x^3 - 3x^2 + 3x - 1) \\ &= (x^4 - 3x^3 + 3x^2 - x) + (x^3 - 3x^2 + 3x - 1) \\ &= x^4 + (-3x^3 + x^3) + (3x^2 - 3x^2) + (-x + 3x) - 1 \\ &= x^4 - 2x^3 + 0x^2 + 2x - 1 \\ &= x^4 - 2x^3 + 2x - 1 \end{align*}

This matches the original polynomial with the coefficients we found. Thus, the other factor is indeed $(x+1)$.

Revision Table: Key Concepts

Additional Information: Polynomial Factorization Techniques

Besides using root properties and derivative tests, other methods to find polynomial factors include:

• Polynomial Long Division: Once we found the polynomial $x^4 - 2x^3 + 2x - 1$ and knew $(x-1)^3 = x^3 - 3x^2 + 3x - 1$ was a factor, we could perform long division to find the quotient, which is the other factor.
• Synthetic Division: If the roots are known (like the triple root at $x=1$), synthetic division can be used repeatedly to reduce the polynomial's degree and find the remaining factors. We could divide $P(x)$ by $(x-1)$ three times.
• Rational Root Theorem: For polynomials with integer coefficients, this theorem helps identify potential rational roots, which can then be tested.

In this specific problem, using the properties of roots and their relation to polynomial coefficients (especially the sum of roots) was an efficient way to find the remaining linear factor after determining the polynomial coefficients $\alpha, \beta, \gamma$.