Problem 25 - Precalculus: Functions and Graphs

Question

Show that the number is a zero of \(f(x)\) of the given multiplicity, and express \(f(x)\) as a product of linear factors. $$f(x)=x^{6}-4 x^{5}+5 x^{4}-5 x^{2}+4 x-1$$1 (multiplicity 5)

Step-by-Step Solution

Answer

The number 1 is a zero with multiplicity 5, and \(f(x) = (x - 1)^5 (x + 1)\).

Step 1: Verify Zero of the Function

Let's evaluate the function \(f(x)\) at \(x = 1\). Substitute \(x = 1\) into \(f(x)\): \(f(1) = 1^6 - 4 \cdot 1^5 + 5 \cdot 1^4 - 5 \cdot 1^2 + 4 \cdot 1 - 1 = 1 - 4 + 5 - 5 + 4 - 1 = 0\). Since \(f(1) = 0\), \(x = 1\) is a zero of \(f(x)\).

Step 2: Synthetic Division (First Iteration)

To confirm the multiplicity, we divide \(f(x)\) by \(x - 1\). Using synthetic division on \(f(x) = x^6 - 4x^5 + 5x^4 - 5x^2 + 4x - 1\) with the divisor \(x - 1\):1 | 1 -4 5 0 -5 4 -1 | 1 -3 2 2 -3 1 |------------------------- 1 -3 2 2 -3 1 0Resulting polynomial is \(x^5 - 3x^4 + 2x^3 + 2x^2 - 3x + 1\). The remainder is 0, confirming \(x = 1\) as a zero.

Step 3: Further Synthetic Divisions

Repeat synthetic division for \(x^5 - 3x^4 + 2x^3 + 2x^2 - 3x + 1\) by \(x - 1\):1 | 1 -3 2 2 -3 1 | 1 -2 0 2 -1 |------------------ 1 -2 0 2 -1 0 Resulting polynomial is \(x^4 - 2x^3 + 0x^2 + 2x - 1\). Since the remainder is 0, continue the process.

Step 4: Further Reduction by Dividing

Continue synthetic division for \(x^4 - 2x^3 + 2x - 1\) by \(x - 1\):1 | 1 -2 0 2 -1 | 1 -1 -1 1 |------------------ 1 -1 -1 1 0 The resulting polynomial is \(x^3 - x^2 - x + 1\). The remainder 0 indicates another successful factorization.

Step 5: Complete Factorization

Continue for \(x^3 - x^2 - x + 1\) by \(x - 1\):1 | 1 -1 -1 1 | 1 0 -1 |------------- 1 0 -1 0Resulting polynomial is \(x^2 - 1\), confirming another zero. \(x^2 - 1 = (x - 1)(x + 1)\).

Step 6: Expressing as Linear Factors

The complete factorization is \(f(x) = (x - 1)^5 (x + 1)\). Here, \((x - 1)^5\) demonstrates that \(x = 1\) has multiplicity 5.

Key Concepts

Synthetic Division

Synthetic division is an efficient method used to divide a polynomial by a linear factor of the form \(x - c\). It's particularly helpful in factoring polynomials and finding polynomial zeros with less work than long division requires.
To perform synthetic division:

• Write the coefficients of the polynomial in order.
• Bring down the leading coefficient.
• Multiply the divisor root by the value brought down, then add this result to the next coefficient.
• Repeat this process until all coefficients have been accounted for.

In this exercise, synthetic division was used repetitively to divide the polynomial \(f(x) = x^6 - 4x^5 + 5x^4 - 5x^2 + 4x - 1\) by \(x - 1\) to confirm the multiplicity of the root and simplify the polynomial into more digestible factors. This process confirmed the zero \(x = 1\) effectively.

Multiplicity of Roots

The multiplicity of a root refers to the number of times a particular root occurs in a polynomial. A root of multiplicity greater than one indicates the zero appears multiple times when the polynomial is factored completely.
In our given exercise, it is noted that \(x = 1\) is a zero of multiplicity 5. This means that when expressing the polynomial \(f(x)\) as a product of its linear factors, \(x - 1\) appears five times. This indicates that the polynomial touches or crosses the x-axis at exactly one point corresponding to \(x = 1\) with a certain repeated behavior as evidenced by its graph. Recognizing the multiplicity of roots is vital when comprehensively grasping the structure and behavior of polynomial functions.

Linear Factors

A polynomial can be expressed as a product of linear factors, which are expressions of degree 1, such as \(x - c\). Linear factors reveal the roots of a polynomial as each factor \(x - c\) corresponds to a root \(c\).
For the polynomial \(f(x)\), the factorization led to the expression \((x-1)^5(x+1)\). Each linear factor represents a solution to the equation \(f(x) = 0\). Here, \(x - 1\) is raised to the fifth power due to its multiplicity, and \(x + 1\) indicates another distinct root. The expression in terms of linear factors provides an immediate understanding of the zeroes and their behaviors in the polynomial.

Zero of a Function

A zero of a function is a value \(x=c\) for which \(f(c) = 0\). This means the graph of the polynomial crosses or touches the x-axis at \(x = c\).
In this exercise, evaluating \(f(x)\) at \(x = 1\) showed that it is indeed a zero since substituting 1 in place of \(x\) yields a result of zero. This is a critical starting point for factoring the polynomial as it confirms that \(x - 1\) is a valid factor of the function. Identifying zeros helps not only in factorization but also in analyzing the graph of the function and understanding its overall behavior.