Problem 61 - A Graphical Approach to Precalculus with Limits

Question

For each polynomial at least one zero is given. Find all others analytically. $$P(x)=x^{3}-2 x^{2}-5 x+6 ; 3$$

Step-by-Step Solution

Answer

Zeros are -2, 1, and 3.

Step 1: Verify the Given Zero

We need to confirm that the given zero, 3, is indeed a root of the polynomial \(P(x) = x^3 - 2x^2 - 5x + 6\). Substitute \(x = 3\) into the polynomial and check if it equals zero. \[P(3) = 3^3 - 2 \times 3^2 - 5 \times 3 + 6 = 27 - 18 - 15 + 6 = 0\]. Indeed, \(x = 3\) is a root.

Step 2: Perform Polynomial Division

Use synthetic division to divide the polynomial \(P(x)\) by \(x-3\) since we know 3 is a root. Perform the division using the coefficients: \(1, -2, -5, 6\).- Bring down the 1.- Multiply 3 (the root) by 1 and add to the second coefficient: \(-2 + 3 \times 1 = 1\).- Multiply 3 by 1 and add to -5: \(-5 + 3 \times 1 = -2\).- Multiply 3 by -2 and add to 6: \(6 + 3 \times (-2) = 0\).The quotient is \(x^2 + x - 2\) and the remainder is 0.

Step 3: Factor the Quotient Polynomial

The resulting quotient from the division is \(x^2 + x - 2\). Now, factor this quadratic polynomial. Find two numbers that multiply to -2 and add to 1. These numbers are 2 and -1. Therefore, \(x^2 + x - 2\) factors into \((x + 2)(x - 1)\).

Step 4: Find All Zeros of the Polynomial

From the factors \((x - 3)(x + 2)(x - 1)\) of the polynomial \(P(x)\), the zeros are found by setting each factor to zero:- \(x - 3 = 0\) gives \(x = 3\).- \(x + 2 = 0\) gives \(x = -2\).- \(x - 1 = 0\) gives \(x = 1\). Thus, the zeros are \(-2, 1, 3\).

Key Concepts

Polynomial Division

Polynomials are mathematical expressions involving variables and coefficients. They can often be broken down to find solutions more easily, one method of doing this is polynomial division. Imagine dividing a polynomial by another polynomial to find what remains or what multiplies to make the original. This is similar to regular number division but involves a bit more algebraic manipulation.

Polynomial division helps simplify these expressions and is crucial for finding zeros or roots of the polynomial. It can be done using two primary methods: long division and synthetic division. Long division works similarly to numerical long division but can be more cumbersome with larger expressions. Synthetic division, however, can greatly simplify the process provided the conditions are met. We applied synthetic division in the solution to efficiently handle the division.

Synthetic Division

Synthetic division is a simplified method of dividing polynomials. It's especially useful when dividing by a linear polynomial, such as using the factor associated with a known zero of the polynomial. Instead of writing out all the variables and terms, you focus on the coefficients of the polynomial.

In our problem, to divide the polynomial \(P(x) = x^3 - 2x^2 - 5x + 6\) by \(x - 3\), we used the coefficients \(1, -2, -5, 6\) and arranged them in a row. The method involved:

• Starting with the leading coefficient on the left.
• Performing a series of multiplications and additions, involving the given zero \(3\).
• Bringing down the leading coefficient, multiplying it by the given zero, and adding it to the next coefficient.

This simplified several steps of polynomial long division into a manageable calculation, culminating in a quotient \(x^2 + x - 2\) with a remainder of zero, confirming \(x - 3\) is a factor.

Factoring Quadratics

Once a quotient polynomial like \(x^2 + x - 2\) is obtained, the next step is to factor it. Factoring finds simpler polynomials that multiply together to give the original. In this way, you uncover potential roots of the polynomial.

For quadratics, you look for two numbers that multiply to the constant term (\(-2\) in this case) and add up to the linear coefficient (\(1\)). Here, the numbers are \(2\) and \(-1\), meaning \(x^2 + x - 2\) factors into \((x + 2)(x - 1)\). This is a key step as it turns a more complex polynomial into simpler binomials, each giving a zero of the polynomial when set equal to zero.

Roots of Polynomials

Finding the roots or zeros of a polynomial is discovering the values of \(x\) that turn the polynomial result to zero. These values are critical in solving polynomial equations as they represent where the graph of the polynomial crosses the x-axis.

Once the polynomial \(P(x)\) is factored completely into \((x - 3)(x + 2)(x - 1)\), finding the zeros is straightforward:

• Set each factor equal to zero and solve for \(x\).
• This yields the zeros: \(x = 3\), \(x = -2\), and \(x = 1\).

These roots tell us a lot about the behavior of the polynomial, including where its values change from positive to negative and its intersections with the x-axis. Recognizing and solving for these roots is essential for understanding polynomial behavior.