Problem 80 - Algebra for College Students

Question

For each pair of polynomials, use division to determine whether the first polynomial is a factor of the second. Use synthetic division when possible. If the first polynomial is a factor, then factor the second polynomial. See Example 7. $$x-1, x^{3}+3 x^{2}-5 x$$

Step-by-Step Solution

Answer

No, \(x-1\) is not a factor of \(x^3 + 3x^2 - 5x\).

Step 1: Set up the synthetic division

The divisor is given as \(x - 1\). For synthetic division, the root used is 1. Write down the coefficients of \(x^3 + 3x^2 - 5x\), which are 1, 3, and -5.

Step 2: Perform synthetic division

Arrange as follows and perform the synthetic division by bringing down the first coefficient, then multiply by 1, add the result to the next coefficient, and continue this process.

Step 3: Conclusion

Since the remainder is not zero, \(x - 1\) is not a factor of \(x^3 + 3x^2 - 5x\), and thus the second polynomial cannot be factored using \(x - 1\).

Key Concepts

synthetic division

Synthetic division is a simplified form of polynomial division, specifically designed for dividing a polynomial by a linear factor of the form \(x - c\). This method reduces the complexity of traditional long division. To perform synthetic division, follow these steps:

1. Identify the root from the divisor \(x - c\), in this case, the root is \(c\).
2. Write down the coefficients of the polynomial you are dividing.
3. Set up a synthetic division layout with the root on the left and the coefficients on the right.
4. Bring down the first coefficient. Multiply it by the root and add the result to the next coefficient. Repeat this for all coefficients.
5. The last number in the row is the remainder.

For our example, the root is 1, and the coefficients of \(x^3 + 3x^2 - 5x\) are 1, 3, and -5:

\[ \begin{array}{r|rrrr}1 & 1 & 3 & -5 & 0 \ \hline & 1 & 4 & -1 & 4 \end{array} \]

Since the remainder is not zero, \(x - 1\) is not a factor of \(x^3 + 3x^2 - 5x\).

polynomial factorization

Polynomial factorization is the process of breaking down a polynomial into simpler polynomials that, when multiplied together, give the original polynomial. This can involve recognizing common factors, using synthetic division, or applying other algebraic techniques.
When a polynomial cannot be factored by simple inspection, synthetic division can help determine if a polynomial has specific factors by breaking it down step by step.
For instance, if a polynomial can be divided by another and the remainder is zero, it indicates factorization. Otherwise, as shown in our example, where \(x - 1\) leaves a remainder, it hints that the respective factor is not valid. This process is vital to simplifying polynomial equations and solving polynomial functions.

remainder theorem

The Remainder Theorem is a useful tool in polynomial division. It states that the remainder of the division of a polynomial \(f(x)\) by a linear divisor \(x - c\) is the same as \(f(c)\). This means that by substituting \(c\) into the polynomial, you can quickly find the remainder.
Using synthetic division to find this remainder is efficient. If the remainder is zero, it confirms that \(x - c\) is a factor of the polynomial. In our case study, substituting 1 into \(f(x) = x^3 + 3x^2 - 5x\) doesn't yield 0, thus verifying via both synthetic division and substitution that \(x - 1\) is not a factor.
The Remainder Theorem not only helps in verifying factors quickly but is integral to understanding the behavior of polynomials and streamlining division processes.