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Magazine Chapter 6: Problem 19
Given a polynomial and one of its factors, find the remaining factors of the polynomial. Some factors may not be binomials. $$ x^{3}-x^{2}-10 x-8 ; x+1 $$
Short Answer
Expert verified
The factors are \((x + 1)(x - 4)(x + 2)\).
Step by step solution
01
Understand the Problem
We are given a polynomial \(x^3 - x^2 - 10x - 8\) and one of its factors \(x + 1\). We need to find all the remaining factors of the polynomial.
02
Use Synthetic Division
We will use synthetic division to divide the polynomial \(x^3 - x^2 - 10x - 8\) by the factor \(x + 1\). First, write down the coefficients \([1, -1, -10, -8]\). Place the zero of \(x + 1\), which is \(-1\), to the left.
03
Perform Synthetic Division
1. Bring down the 1.2. Multiply -1 by 1 (get -1), place under the next coefficient (-1).3. Add to -1 to get -2.4. Multiply -1 by -2 (get 2), place under next coefficient (-10).5. Add to -10 to get -8.6. Multiply -1 by -8 (get 8), place under -8.7. Add to -8 to get 0. Thus, the quotient is \(x^2 - 2x - 8\) and remainder is 0, confirming \(x + 1\) is a factor.
04
Factor the Quotient Polynomial
Now, factor \(x^2 - 2x - 8\). First, find two numbers that multiply to -8 and add to -2. They are -4 and 2. Hence, \(x^2 - 2x - 8 = (x - 4)(x + 2)\).
05
Write All Factors
Since \(x^3 - x^2 - 10x - 8\) was divided by \(x + 1\) to get \(x^2 - 2x - 8 = (x - 4)(x + 2)\), the complete factorization of the polynomial is \((x + 1)(x - 4)(x + 2)\).
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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 simplified way to divide polynomials, particularly when the divisor is a linear binomial like \(x + 1\). This method streamlines the division process by focusing on the coefficients rather than the full algebraic expressions, which is the key difference from polynomial long division. To illustrate:
- Identify the zero of the divisor, which in this case is \(-1\) from the factor \(x + 1\).
- Write down the coefficients of the polynomial \([1, -1, -10, -8]\).
- Carry out the synthetic division process by multiplying and adding as previously outlined, ultimately confirming the factor division is exact and yielding a remainder of 0.
Polynomial Long Division
Polynomial long division is akin to the standard long division that one might use with numbers but adapted to handle polynomials. It is a step-by-step method where we divide the larger polynomial by another polynomial, generally of a lower degree.Imagine you're dividing \(x^3 - x^2 - 10x - 8\) by \(x + 1\):
- The first term \(x^3\) is divided by the leading term of the divisor \(x\), resulting in \(x^2\).
- Multiply \(x^2\) by \(x + 1\) and subtract the result from the original polynomial.
- Repeat the process with the new polynomial until the degree of the remainder is less than the degree of the divisor.
Factoring Quadratics
Factoring quadratics, such as the expression \(x^2 - 2x - 8\), involves breaking down the expression into simpler factors. The aim is to find two binomials whose product gives the original quadratic.To factor \(x^2 - 2x - 8\):
- Look for two numbers that multiply to \(-8\) (the constant term) and add to \(-2\) (the coefficient of \(x\)).
- In this instance, \(-4\) and \(2\) meet the criteria as \((-4) \times 2 = -8\) and \((-4) + 2 = -2\).
- This results in the binomial factors \((x - 4)(x + 2)\).
Remainder Theorem
The Remainder Theorem is a useful tool that links division and function evaluation. It states that for any polynomial \(f(x)\), the remainder of the division of \(f(x)\) by \((x - c)\) is equal to \(f(c)\).For the polynomial \(x^3 - x^2 - 10x - 8\) with the factor \(x + 1\), the substitute value is \(-1\). So evaluating the polynomial at \(-1\), \(f(-1)\), would yield the remainder:
- Plug in \(-1\) to the polynomial:
- \(f(-1) = (-1)^3 - (-1)^2 - 10(-1) - 8 = -1 - 1 + 10 - 8 = 0\).
- The zero remainder confirms \(x + 1\) is indeed a factor.
