Problem 76
Question
Use the given zero to completely factor \(P(x)\) into linear factors. $$\text { Zero: } 2-i ; \quad P(x)=x^{4}-4 x^{3}+9 x^{2}-16 x+20$$
Step-by-Step Solution
Verified Answer
The factorization is \((x-(2-i))(x-(2+i))(x-2)^2\).
1Step 1: 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 \(x - (2 + i)\).
2Step 2: 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:\[(x - 2)^2 + 1 = x^2 - 4x + 5.\]
3Step 3: 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 \(Q(x)\).
4Step 4: 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:\[x^2 - 4x + 4 = (x - 2)^2.\]
5Step 5: Write the complete factorization
Combine the quadratic factor and its resulting linear factors to express \(P(x)\) completely:\[P(x) = (x - 2 + i)(x - 2 - i)(x - 2)(x - 2) \text{ or } (x - (2 - i))(x - (2 + i))(x - 2)^2.\]
Key Concepts
Complex NumbersPolynomial DivisionConjugate PairsLinearity
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.
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 \((x - 2)^2\).
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 \((x - (2 - i))(x - (2 + i))\).
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 \(2\).
Other exercises in this chapter
Problem 75
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Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=-6 x^{3}-17 x^{2}+63 x-10 ; \quad k=-5$$
View solution Problem 76
Simplify each power of i to \(i, 1,-i,\) or \(-1\). $$i^{19}$$
View solution