Problem 24 - 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. $$f(x)=x^{4}-9 x^{3}+22 x^{2}-32 ; \quad 4 \text { (multiplicity 2) }$$

Step-by-Step Solution

Answer

\(f(x) = (x-4)^2(x-2)(x+1)\).

Step 1: Verify if 4 is a Zero of f(x)

To verify that 4 is a zero of the polynomial \(f(x) = x^4 - 9x^3 + 22x^2 - 32\), substitute \(x = 4\) into the polynomial: \[ f(4) = (4)^4 - 9(4)^3 + 22(4)^2 - 32. \]Calculate each term:- \( (4)^4 = 256 \),- \( 9(4)^3 = 9 imes 64 = 576 \),- \( 22(4)^2 = 22 imes 16 = 352 \),- Subtraction of 32.Combine terms: \( 256 - 576 + 352 - 32 = 0 \). Thus, \(4\) is indeed a zero of \(f(x)\).

Step 2: Use Synthetic Division to Find Quotient

Since \(4\) is a zero, we divide \(f(x)\) by \(x-4\) using synthetic division. The coefficients are \(1, -9, 22, 0, -32\). Proceed as follows:1. Bring down the leading coefficient (1).2. Multiply by 4: \(1 \cdot 4 = 4\), add to next coefficient: \(-9 + 4 = -5\).3. Repeat: \(-5 \cdot 4 = -20\), add to next coefficient: \(22 - 20 = 2\).4. Repeat: \(2 \cdot 4 = 8\), add to next coefficient (0): \(0 + 8 = 8\).5. Repeat: \(8 \cdot 4 = 32\), add to last term: \(-32 + 32 = 0\).The quotient is \(x^3 - 5x^2 + 2x + 8\). The remainder is 0, confirming \(x-4\) is a factor.

Step 3: Confirm Multiplicity with Further Division

Now confirm the multiplicity by dividing the resultant polynomial \(x^3 - 5x^2 + 2x + 8\) by \(x-4\) again using synthetic division:1. Bring down the leading coefficient (1).2. Multiply by 4: \(1 \cdot 4 = 4\), add to next coefficient: \(-5 + 4 = -1\).3. Repeat: \(-1 \cdot 4 = -4\), add to next coefficient: \(2 - 4 = -2\).4. Repeat: \(-2 \cdot 4 = -8\), add to last term: \(8 - 8 = 0\).The quotient is \(x^2 - x - 2\) with a remainder of 0, confirming \((x-4)^2\) as a factor.

Step 4: Factor the Remaining Quadratic

Factor the quadratic \(x^2 - x - 2\) by looking for two numbers that multiply to -2 and add to -1: These numbers are -2 and 1. The quadratic factors as \((x - 2)(x + 1)\).

Step 5: Express f(x) as a Product of Linear Factors

Combine the results from the previous steps. Since 4 is a zero of multiplicity 2, \(x-4\) is squared, and you have factored \(x^2 - x - 2\) as \((x - 2)(x + 1)\), express \(f(x)\) as:\[ f(x) = (x - 4)^2 (x - 2)(x + 1) \].

Key Concepts

Polynomial Zeros

One of the fundamental concepts in understanding polynomials is the idea of polynomial zeros. A zero of a polynomial is simply a solution to the equation when the polynomial is set equal to zero. In other words, if you replace the variable (usually represented as \(x\)) with a specific number and the result is zero, then that number is called a zero of the polynomial. This zero is where the graph of the polynomial will cross the x-axis.

In our example, we confirmed that 4 is a zero of the polynomial function \(f(x) = x^4 - 9x^3 + 22x^2 - 32\). By substituting \(x = 4\) into the function and simplifying, we obtained a result of zero, validating that 4 is indeed a zero of the polynomial.

Knowing the zeros of a polynomial is crucial for factorization and understanding the behavior of the polynomial function.

Synthetic Division

Synthetic division is a shortcut method for dividing a polynomial by a linear factor of the form \(x - c\). It is a simplified alternative to long division and is particularly useful when testing for zeros.

Here's how synthetic division works:

• Write down the coefficients of the polynomial.
• Bring down the leading coefficient to the bottom row.
• Multiply this number by the zero you are testing (in our case, 4) and write the result under the next coefficient.
• Add the column, repeat the process.

If the final row ends with a zero, \(x - c\) is a factor!

In our exercise, using synthetic division, we confirmed that \(x-4\) is a factor of \(f(x)\) and found a quotient of \(x^3 - 5x^2 + 2x + 8\). The remainder was zero, proving our zero and factor were correct.

Linear Factors

Linear factors are expressions of the form \(x - c\) where \(c\) is a constant. Expressing a polynomial as a product of its linear factors is a key step in factorization. It essentially breaks down a polynomial into simpler, root-specific components.

For the polynomial \(f(x) = x^4 - 9x^3 + 22x^2 - 32\), we broke it down into its linear factors by repeatedly dividing by known zeros. First, we divided by \(x-4\) twice because 4 has a multiplicity of 2. Further factorization of the remaining polynomial \(x^2 - x - 2\) yielded the linear factors \((x-2)\) and \((x+1)\).

This results in the expression \(f(x) = (x-4)^2(x-2)(x+1)\), showing the polynomial as a product of its linear factors.

Zero Multiplicity

Zero multiplicity refers to the number of times a particular zero occurs for a given polynomial. In essence, it's the number of times that \(x - c\) appears as a factor.Multiplicity has important implications for the shape of a polynomial's graph. Zeros of odd multiplicity, for example, cross the x-axis, while those of even multiplicity do not; they just touch the axis and turn back.

In the exercise above, the zero 4 has a multiplicity of 2, meaning \(x-4\) appears twice in the factorization of the polynomial as \((x-4)^2\). This indicates that the graph of \(f(x)\) touches the x-axis at \(x=4\) without crossing it, creating a "bounce." Understanding multiplicity enriches our grasp of polynomial behavior and factorization.