Problem 19 - Algebra 2

Question

Given a polynomial and one of its factors, find the remaining factors of the polynomial. Some factors may not be binomials. $$ x^{3}-x^{2}-10 x-8 ; x+1 $$

Step-by-Step Solution

Answer

The factors are \((x + 1)(x - 4)(x + 2)\).

Step 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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, \(x^2 - 2x - 8 = (x - 4)(x + 2)\).

Step 5: 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 \((x + 1)(x - 4)(x + 2)\).

Key Concepts

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.

This approach efficiently reduces the polynomial to a simpler form, making further factorization easier.

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.

In our case, using synthetic division, which complements polynomial long division, was more efficient, leading us directly to \(x^2 - 2x - 8\) as the quotient.

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 \((x - 4)(x + 2)\).

Mastering this skill is crucial for working with polynomials, as it helps in simplifying expressions and solving equations.

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.

This theorem simplifies the factor verification process, especially when combined with synthetic division or long division.