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Chapter 19: Problem 12

Factor \(x^{5}-1\) into linear and real quadratic factors.

Short Answer

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Question: Factor the polynomial \(x^5 - 1\) into linear and real quadratic factors. Answer: \((x - 1)(x^2 + x + 1)(x^2 - x + 1)\)

Step by step solution

01

Use the difference of odd powers rule

Using the difference of odd powers rule, we can factor \(x^5 - 1\) as: \((x^5 - 1) = (x - 1)(x^4 + x^3 + x^2 + x + 1)\).
02

Apply the quadratic factorization method for \(x^4 + x^3 + x^2 + x + 1\)

To break down the expression further, we'll look for any complex conjugate pairs in the factors of \(x^4 + x^3 + x^2 + x + 1\). Let \(z = x^2 + bx + c\) be a quadratic factor of \(x^4 + x^3 + x^2 + x + 1\). We know that for each linear factor \(r(x)\) of \(z\), its complex conjugate \(r^*(x)\) will also be a factor. So, if \(z(x) = (ax + p)(ax + q)\) where \(p\) and \(q\) are complex conjugates, then \(z^*(x) = (ax + p^*)(ax + q^*)\). Using the observation that the product of a complex number and its conjugate results in a real number: \((x^2 + bx + c)(x^2 - bx + c) = x^4 + x^3 + x^2 + x + 1\). Now, let's multiply the expressions and compare coefficients: \((x^2 + bx + c)(x^2 - bx + c) = x^4 + (c^2 - b^2)x^2 + c^2 x^2 + c^2 x_-_ c^2 x + c^2\). Comparing the coefficients, we get: 1. \(c^2 - b^2 = 1\) 2. \(c^2 = 1\) 3. \(c^2 = 1\) Solving for \(c\), we find that \(c = \pm1\).
03

Solve for \(b\)

Since we now know \(c = 1\), we can substitute it into the first coefficient comparison expression to solve for \(b\): \(1 - b^2 = 1\) \(b^2 = 0\) \(b = 0\) Now, we can rewrite the quadratic factors with the determined values for \(b\) and \(c\): \(x^4 + x^3 + x^2 + x + 1 = (x^2 + x + 1)(x^2 - x + 1)\).
04

Write the final factored polynomial

Combining the factored expressions from Steps 1 and 3, we have: \(x^5 - 1 = (x - 1)(x^2 + x + 1)(x^2 - x + 1)\). The final factored polynomial is \((x - 1)(x^2 + x + 1)(x^2 - x + 1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Odd Powers
When factoring an expression like \(x^5 - 1\), the difference of odd powers rule is a valuable tool. This rule helps us break down such expressions into simpler factors. The rule states that for any odd power \(n\), the expression \(x^n - a\) can be factored as:
  • \((x - a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + a^{n-1})\)
Applying this rule to \(x^5 - 1\), consider \(a = 1\). So, we get the factorization:
  • \((x - 1)(x^4 + x^3 + x^2 + x + 1)\)
The first factor \((x - 1)\) is linear, and the second part \((x^4 + x^3 + x^2 + x + 1)\) is the sum of powers part. This makes for a great starting point for further simplification in the factorization process.
Complex Conjugate Pairs
Complex conjugate pairs come into play notably when dealing with polynomials with non-real roots. In our factorization problem, the expression \(x^4 + x^3 + x^2 + x + 1\) can be potentially broken down into quadratic factors that involve complex roots.
For any pair of complex numbers \(p\) and \(\overline{p}\) (its conjugate), the product is a real number. This is important for transforming complex expressions into real factors.
Considering this for our expression, we check for such pairs within quadratic factors of \(x^4 + x^3 + x^2 + x + 1\). We identify two quadratic factors:
  • \((x^2 + x + 1)\)
  • \((x^2 - x + 1)\)
Each of these can be simplified using their respective complex roots. Both pairs can be demonstrated to produce real quadratic factors that contribute to the overall factorization.
Quadratic Factorization
Quadratic factorization further simplifies polynomial expressions to make them easier to handle or solve. From our earlier step, after acknowledging the pattern in \(x^4 + x^3 + x^2 + x + 1\), we utilize quadratic factorization by solving for potential quadratic components.
We explore possible combinations that maintain the balance in coefficients, leading us to confirm:
  • \((x^2 + x + 1)\)
  • \((x^2 - x + 1)\)
Solving these gives us the consistent set of quadratic factors. We ensure these are correct by verifying product expansions to compare coefficients with the original expression. This is a common practice that helps double-check factorization accuracy.
Thus, via quadratic factorization, the original polynomial \(x^5 - 1\) becomes the product \((x-1)(x^2+x+1)(x^2-x+1)\), which brings it into complete factorized form using both linear and real quadratic components.

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Most popular questions from this chapter

Nicolaus Bernoulli claimed that the polynoma1 \(x^{2}-4 x=\) \(2 x^{2}+4 x+4\) could not be factored into a product of two quadratic polynomials. Show that in fact the factors are $$ \left(x^{2}-(2+\sqrt{4+2 \sqrt{7}}) x+(1+\sqrt{4+2 \sqrt{7}}+\sqrt{7})\right) $$ and $$ \left(x^{2}-(2-\sqrt{4+2 \sqrt{7}}) x+(1-\sqrt{4+2 \sqrt{7}}+\sqrt{7})\right) $$.

A horse dealer bought a horse for a certain number of crowns and sold it again for 119 crowns, by which means his profit was as much per cent as the horse cost him. What was the purchase price?

Of three workmen, \(A\) can finish a given job once in three weeks, \(B\) can finish it three times in eight weeks, while \(C\) can finish it five times in twelve weeks. How long will it take for the three workmen to complete the job together? (This exercise and the next two are from Newton's Universal Arithmetic.)

Benjamin Banneker was fond of solving mathematical puzzles and recorded many in his notebook, including his own version of the old hundred fowls problem: A gentleman sent his servant with \(拢 100\) to buy 100 cattle, with orders to give \(拢 5\) for each bullock, 20 shillings for each cow, and 1 shilling for each sheep. (Recall that 20 shillings equals \(拢 1\).) What number of each sort of animal did he bring back to his master? \(?^{23}\)

Show that if \(n\) is prime, then the roots of \(x^{n}-1=0\) can all be expressed as powers of any such root \(\alpha \neq 1\).
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