Chapter 4: Problem 73
Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=6 x^{3}+25 x^{2}+3 x-4 ; \quad k=-4$$
Short Answer
Expert verified
The polynomial is factored as \((x + 4)(2x + 1)(3x - 1)\).
Step by step solution
01
Synthetic Division Setup
To factor the polynomial given that \( k = -4 \) is a zero, we will use synthetic division. Write \( k = -4 \) to the left and align the coefficients of \( P(x) = 6x^3 + 25x^2 + 3x - 4 \) as a row: [6, 25, 3, -4].
02
Perform Synthetic Division
Start by bringing down the leading coefficient, 6. Multiply it by \( k = -4 \) and add the result to the next coefficient: \( (6 \times -4) + 25 = 1 \). Repeat this process: \( (1 \times -4) + 3 = -1 \) and then \( (-1 \times -4) + -4 = 0 \). Our new row becomes [6, 1, -1, 0].
03
Identify the Quotient Polynomial
From synthetic division, the polynomial can be written as a product of \((x + 4)\) and \(6x^2 + x - 1\).
04
Factor the Quadratic Term
Now factor the quadratic term \( 6x^2 + x - 1 \). We need factors of \( -6 \) that add to 1, which are 3 and -2. Rewrite the middle term: \( 6x^2 + 3x - 2x - 1 \). Factor by grouping: \((6x^2 + 3x) - (2x + 1)\) which becomes \(3x(2x + 1) - 1(2x + 1)\). Factor out \((2x + 1)\), resulting in \((2x + 1)(3x - 1)\).
05
Write the Complete Factorization
Combine all factors: The complete factorization of \( P(x) \) is \((x + 4)(2x + 1)(3x - 1)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Synthetic Division
Synthetic division is a method used to divide a polynomial by a binomial of the form \( x - k \). It's more efficient than long division, especially when dealing with coefficients and determining a polynomial's factors. Let's break down how it works:
- First, identify the zero \( k \) of the polynomial. In our exercise, this is \( k = -4 \).
- Next, write down the coefficients of the polynomial in a row. For the polynomial \( 6x^3 + 25x^2 + 3x - 4 \), these numbers are [6, 25, 3, -4].
- Perform operations step by step, bringing down the initial coefficient, multiplying by the zero \( k \), and adding the result to the next coefficient.
The final results give you both the quotient polynomial and the remainder. If the remainder is zero, as in our case, \( k \) is indeed a zero of the polynomial.
- First, identify the zero \( k \) of the polynomial. In our exercise, this is \( k = -4 \).
- Next, write down the coefficients of the polynomial in a row. For the polynomial \( 6x^3 + 25x^2 + 3x - 4 \), these numbers are [6, 25, 3, -4].
- Perform operations step by step, bringing down the initial coefficient, multiplying by the zero \( k \), and adding the result to the next coefficient.
The final results give you both the quotient polynomial and the remainder. If the remainder is zero, as in our case, \( k \) is indeed a zero of the polynomial.
Zero of a Polynomial
A zero of a polynomial, also known as a root, is a value for which the polynomial is equal to zero. When you replace \( x \) with this specific zero, the polynomial's output should be zero.
- In our specific example, we are given that \( k = -4 \) is a zero of the polynomial \( P(x) = 6x^3 + 25x^2 + 3x - 4 \).
- Finding such zeros is crucial because it helps us break down a polynomial into simpler parts (i.e., factor it).
- Knowing a zero allows us to factor the polynomial, which can lead to finding all the polynomial's zeros.
Quadratic Factoring
Quadratic factoring involves breaking down a quadratic expression into two binomials. It's an essential part of polynomial factorization once we have reduced the degree of our polynomial using methods like synthetic division.
- The reduced polynomial from synthetic division in our example was \( 6x^2 + x - 1 \).
- To factor this, find two numbers that multiply to the product of the leading coefficient (6) and the constant term (-1), and also add up to the middle coefficient (1).
- The numbers are 3 and -2. So, split the middle term and perform factoring by grouping.
- This results in \( (3x - 1)(2x + 1) \).
Polynomial Zeros
Once a polynomial is factored completely, determining the zeros becomes straightforward. The zeros are the values of \( x \) that make each factor equal to zero.
- From our original exercise, the polynomial \( P(x) \) factors into \( (x + 4)(2x + 1)(3x - 1) \).
- Set each factor equal to zero: \( x + 4 = 0 \), \( 2x + 1 = 0 \), and \( 3x - 1 = 0 \).
- Solving these, we find the zeros are \( x = -4 \), \( x = -\frac{1}{2} \), and \( x = \frac{1}{3} \).
Polynomial zeros are vital for understanding the behavior of polynomial functions, such as where the graph intersects the x-axis. They offer insights into the function's structure and solutions.
- From our original exercise, the polynomial \( P(x) \) factors into \( (x + 4)(2x + 1)(3x - 1) \).
- Set each factor equal to zero: \( x + 4 = 0 \), \( 2x + 1 = 0 \), and \( 3x - 1 = 0 \).
- Solving these, we find the zeros are \( x = -4 \), \( x = -\frac{1}{2} \), and \( x = \frac{1}{3} \).
Polynomial zeros are vital for understanding the behavior of polynomial functions, such as where the graph intersects the x-axis. They offer insights into the function's structure and solutions.