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Chapter 3: Problem 76

Use the given zero to completely factor \(P(x)\) into linear factors. $$\text { Zero: } 2-i ; \quad P(x)=x^{4}-4 x^{3}+9 x^{2}-16 x+20$$

Short Answer

Expert verified
The factorization is \((x-(2-i))(x-(2+i))(x-2)^2\).

Step by step solution

01

Use the given complex zero

If a polynomial has a complex zero, its conjugate will also be a zero. Given the zero is \(2 - i\), its conjugate \(2 + i\) is also a zero of \(P(x)\). This gives us two factors: \(x - (2 - i)\) and \(x - (2 + i)\).
02

Form a quadratic factor

Multiply the factors \((x - (2 - i))(x - (2 + i))\). Using the difference of squares formula, this becomes:\[(x - 2 + i)(x - 2 - i) = (x - 2)^2 - i^2 = (x - 2)^2 + 1.\]This simplifies to:\[(x - 2)^2 + 1 = x^2 - 4x + 5.\]
03

Perform polynomial division

The quadratic \(x^2 - 4x + 5\) is a factor of \(P(x)\). Perform polynomial division of \(P(x)\) by \(x^2 - 4x + 5\) to find the other factor. This yields another quadratic factor. Let's call the quotient \(Q(x)\).
04

Factor the quotient

After performing the division, suppose we get the quotient polynomial as \(Q(x) = x^2 - 4x + 4\). Notice that this can be factored further as:\[x^2 - 4x + 4 = (x - 2)^2.\]
05

Write the complete factorization

Combine the quadratic factor and its resulting linear factors to express \(P(x)\) completely:\[P(x) = (x - 2 + i)(x - 2 - i)(x - 2)(x - 2) \text{ or } (x - (2 - i))(x - (2 + i))(x - 2)^2.\]

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Key Concepts

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

Complex Numbers
Complex numbers are numbers that have both a real part and an imaginary part. They are written in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\). In the context of polynomial factorization, complex numbers play a crucial role because they can be solutions (roots) of polynomials that do not have real solutions.
  • For example, if we have a polynomial with a root of \(2 - i\), breaking it down requires understanding both the real (\(2\)) and imaginary components (\(-i\)).
  • This ability to solve equations via complex numbers expands our ability to work with different polynomial types beyond simple real number solutions.
Understanding complex numbers allows us to factorize polynomials that involve square roots of negative numbers, which is impossible when using only real numbers.
Polynomial Division
Polynomial division is a method used to determine other factors of a polynomial when one factor is already known. This is very similar to regular numerical division, but applied to polynomials. Given a polynomial and a divisor polynomial, we divide to find a quotient and a remainder. In cases of complete factorization, the remainder should equal zero.
  • In our example, the division of \(P(x) = x^4 - 4x^3 + 9x^2 - 16x + 20\) by the known factor \(x^2 - 4x + 5\) allows us to simplify further and find more factors.
  • The quotient can often be factored further, as seen with \(Q(x) = x^2 - 4x + 4\), which is \((x - 2)^2\).
This process requires careful alignment and subtraction, similar to long division in arithmetic. It's crucial for breaking down higher degree polynomials into manageable, simpler polynomial factors.
Conjugate Pairs
Conjugate pairs are very important in polynomial factorization, especially when dealing with complex numbers. When a polynomial has a complex root, its conjugate is also a root. The conjugate of a complex number \(a + bi\) is \(a - bi\). This is a mathematical property that assists in simplifying polynomials with complex roots.
  • For example, if \(2 - i\) is a root, then its conjugate \(2 + i\) must also be a root of the polynomial.
  • This means when finding factors, you automatically gain another factor related to the conjugate, which aids in building quadratic expressions as shown with \((x - (2 - i))(x - (2 + i))\).
Conjugate pairs ensure that when multiplying these factors together, the imaginary components disappear, yielding a real quadratic or linear factor, simplifying the factorization process.
Linearity
Linearity in this context refers to expressing a polynomial in terms of its linear factors. A linear factor is an expression of the form \(x - r\), where \(r\) is a root of the polynomial. When a polynomial is completely factored into linear factors, it shows all possible real and complex roots.
  • For example, the polynomial \(P(x) = (x - (2 - i))(x - (2 + i))(x - 2)^2\) is broken down into its simplest form through linearity.
  • This representation allows us to quickly identify all roots of the polynomial: \(2 - i\), \(2 + i\), and \(2\).
Linearity is crucial for understanding the fundamental theorem of algebra, which asserts that a polynomial of degree \(n\) can be factored completely into \(n\) linear factors if complex numbers are included. This structure helps in analyzing and solving polynomial equations efficiently.

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Most popular questions from this chapter

Use the given zero to completely factor \(P(x)\) into linear factors. $$\text { Zero: } i ; P(x)=x^{5}-x^{4}+5 x^{3}-5 x^{2}+4 x-4$$

For each polynomial function, (a) list all possible rational zeros, (b) use a graph to eliminate some of the possible zeros listed in part (a), (c) find all rational zeros, and (d) factor \(P(x)\). $$P(x)=x^{3}+6 x^{2}-x-30$$

Use the concepts of this section. Show analytically that \(-1\) is a zero of multiplicity 3 of \(P(x)=x^{5}+9 x^{4}+33 x^{3}+55 x^{2}+42 x+12,\) and find all complex zeros. Then, write \(P(x)\) in factored form.

Find all rational zeros of each polynomial function. $$P(x)=\frac{1}{6} x^{4}-\frac{11}{12} x^{3}+\frac{7}{6} x^{2}-\frac{11}{12} x+1$$

Divide. $$\left(x^{3}-x^{2}+1\right) \div\left(2 x^{2}-1\right)$$
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