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Magazine Chapter 4: Problem 7
A polynomial P is given. (a) Find all zeros of P, real and complex. (b) Factor P completely. \(P(x)=x^{3}-2 x^{2}+2 x\)
Short Answer
Expert verified
Zeros: \(x = 0, 1+i, 1-i\). Factored: \(x(x-(1+i))(x-(1-i))\).
Step by step solution
01
Identify Possible Zeros with Rational Root Theorem
The Rational Root Theorem suggests that possible rational zeros of a polynomial \(P(x)=x^{3}-2x^{2}+2x\) are factors of the constant term divided by factors of the leading coefficient. However, in this case, there is no constant to consider and the factor is simply 0, making 0 a candidate.
02
Evaluate Polynomial at Suspected Zeros
Substitute \(x = 0\) into \(P(x) = x^3 - 2x^2 + 2x\). \(P(0) = 0^3 - 2 \, 路 \, 0^2 + 2 \, 路 \, 0 = 0\). Thus, \(x = 0\) is a zero.
03
Divide Polynomial by (x - 0)
Since \(x = 0\) is a zero, perform polynomial division of \(P(x)\) by \(x\). Factoring out \(x\), we get \(P(x) = x(x^2 - 2x + 2)\).
04
Find Remaining Zeros of Quadratic Factor
Solve the quadratic equation \(x^2 - 2x + 2 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = -2, c = 2\). Calculate the discriminant: \((-2)^2 - 4 \, 路 \, 1 \, 路 \, 2 = 4 - 8 = -4\). The discriminant is negative, indicating complex roots.
05
Find Complex Zeros
Using the quadratic formula \(x = \frac{-(-2) \pm \sqrt{-4}}{2 \, 路 \, 1}\), simplify to \(x = \frac{2 \pm \sqrt{-4}}{2}\). This simplifies further to \(x = 1 \pm i\). Thus, the complex zeros are \(x = 1 + i\) and \(x = 1 - i\).
06
Factor the Polynomial Completely
The polynomial \(P(x) = x^3 - 2x^2 + 2x\) can be expressed in terms of its zeros: \(P(x) = x(x - (1+i))(x - (1-i))\). This is its complete factorization involving real and complex factors.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Root Theorem
The Rational Root Theorem is a useful tool when determining possible rational zeros of a polynomial. It states that if there is a rational zero \( \frac{p}{q} \) of a polynomial \( P(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0 \), where all coefficients \( a_i \) are integers, then \( p \) (the numerator) must be a factor of the constant term \( a_0 \), and \( q \) (the denominator) must be a factor of the leading coefficient \( a_n \). Here are some key points:
- The theorem only suggests potential rational roots, not guarantees.
- The practical use is to test a finite number of potential zeros.
- Zeros need to be checked by direct substitution.
Polynomial Division
After discovering a zero of a polynomial, you can perform polynomial division to simplify the expression and find remaining factors. When you know a zero, say \( x = c \), you can factor \( P(x) \) as \( (x - c)Q(x) \), where \( Q(x) \) is the quotient obtained from polynomial division.
- Start by dividing the polynomial by \( (x - c) \).
- Each step involves dividing the leading term of the dividend by the leading term of the divisor.
- Multiply and subtract to find the remainder and continue until all terms are divided.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here is how it works:
- Determine the coefficients \( a \), \( b \), and \( c \) from your equation.
- Compute the discriminant \( b^2 - 4ac \). This tells you the nature of the roots:
- If positive, there are two distinct real roots.
- If zero, there is one real root (a repeated root).
- If negative, the roots are complex (involving imaginary numbers).
Complex Numbers
Complex numbers extend the concept of the one-dimensional number line to a two-dimensional plane by introducing an imaginary unit \( i \), which is defined as \( \sqrt{-1} \). A complex number usually has the form \( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part.
- They are used when no real solution exists for an equation, typically when the discriminant of a quadratic is negative.
- Addition and subtraction of complex numbers act like combining like terms: \((a + bi) + (c + di) = (a + c) + (b + d)i\).
- Multiplying involves using \( i^2 = -1 \): \((a + bi)(c + di) = ac + bci + adi + bdi^2\).
