Problem 66 - A Graphical Approach to College Algebra

Question

For each polynomial, at least one zero is given. Find all others analytically. $$P(x)=x^{3}-7 x^{2}+13 x-3 ; 3$$

Step-by-Step Solution

Answer

The zeros are 3, \(2 + \sqrt{3}\), and \(2 - \sqrt{3}\).

Step 1: Understand the Given Information

The polynomial provided is \(P(x) = x^3 - 7x^2 + 13x - 3\). We are given that one zero (root) of this polynomial is \(x = 3\). Our task is to find all zeros of the polynomial.

Step 2: Use the Given Zero to Factor the Polynomial

Since \(x = 3\) is a zero, \((x - 3)\) is a factor of \(P(x)\). We need to perform polynomial division to factor the polynomial. Divide \(P(x)\) by \(x - 3\) using synthetic division.

Step 3: Perform Synthetic Division

For synthetic division, set up the coefficients of the polynomial: \(1, -7, 13, -3\). Use \(3\) for the division. Perform the division: - Bring down the first coefficient (which is 1).- Multiply 1 by 3, add to the next coefficient (-7) to get -4.- Multiply -4 by 3, add to the next coefficient (13) to get 1.- Multiply 1 by 3, add to the last coefficient (-3) to get 0 (confirming it is a factor).This gives us the quotient: \(x^2 - 4x + 1\).

Step 4: Solve the Quadratic Equation

Now, solve the quadratic polynomial \(x^2 - 4x + 1 = 0\). This can be done using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=1\), \(b=-4\), and \(c=1\).

Step 5: Apply the Quadratic Formula

Compute the discriminant: \((-4)^2 - 4 \times 1 \times 1 = 16 - 4 = 12\). Now apply the formula:\[x = \frac{4 \pm \sqrt{12}}{2}\]Simplify \(\sqrt{12}\) to \(2\sqrt{3}\), and continue:\[x = \frac{4 \pm 2\sqrt{3}}{2}\]Simplify further:\[x = 2 \pm \sqrt{3}\]

Step 6: List All Zeros

The solutions to the quadratic equation are \(x = 2 + \sqrt{3}\) and \(x = 2 - \sqrt{3}\). Together with the given zero \(x = 3\), all zeros of the polynomial are \(3, 2 + \sqrt{3}, 2 - \sqrt{3}\).

Key Concepts

Quadratic Formula

The quadratic formula is a fundamental tool in algebra for solving equations of the form \( ax^2 + bx + c = 0 \). It expresses the solutions directly in terms of the coefficients \( a \), \( b \), and \( c \). The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here’s a breakdown of each component:

• \( b^2 - 4ac \) is known as the discriminant. It determines the nature of the roots.
• The symbol \( \pm \) indicates there are generally two solutions: one involving addition, the other subtraction.
• The term \( 2a \) in the denominator ensures the entire expression is properly scaled.

This formula is particularly useful when factoring is complex or not possible. In our original problem, we used it to find the remaining roots after dividing the cubic polynomial. By substituting \( a=1 \), \( b=-4 \), and \( c=1 \), we calculated the discriminant and found the two other roots: \( x = 2 + \sqrt{3} \) and \( x = 2 - \sqrt{3} \). It's a reliable method because it's applicable to any quadratic equation, providing exact solutions every time.

Synthetic Division

Synthetic division is a simplified method of dividing polynomials, particularly when dividing by a linear factor like \( x - r \). This technique is efficient and involves fewer steps than traditional long division. Synthetic division primarily deals with the coefficients of the polynomial.Here's how synthetic division works in our example:

• We used the known zero \( x = 3 \) of the polynomial \( P(x) = x^3 - 7x^2 + 13x - 3 \). This implies \( x - 3 \) is a factor.
• We listed the coefficients: \( 1, -7, 13, -3 \), and performed operations using \( 3 \).
• In the process, we verified the division by confirming the remainder is zero, proving \( x - 3 \) is indeed a factor.
• Synthetic division left us with a quotient polynomial: \( x^2 - 4x + 1 \).

It simplifies the process by handling only numbers without variable notation, making it quicker for students once they understand the basic principles. It also reinforced the connection between zeros and factors, as evident in the exercise’s goal.

Factoring Polynomials

Factoring polynomials is the process of breaking a polynomial into simpler components whose product is equal to the original polynomial. These components are called "factors" and can be linear or quadratic expressions.For the exercise, once we knew one zero of the polynomial was \( x = 3 \), we were able to factor the expression \( x - 3 \) from the original polynomial through synthetic division.Here are the basic steps when factoring polynomials:

• Identify any known zeros or roots: these give clues about potential factors.
• Perform division (synthetic division in our case) to simplify the polynomial.
• Factor the resulting polynomial further, if possible.
• Verify your factors by multiplying them back to ensure they result in the original polynomial.

Factoring was central to solving the exercise efficiently, as once the polynomial was reduced to \( x^2 - 4x + 1 \), we could apply the quadratic formula to find the remaining roots. Factoring not only simplifies the problem but also provides a path to solutions when combined with methods like the quadratic formula.