Problem 64 - Algebra and Trigonometry

Question

A polynomial P is given. (a) Factor P into linear and irreducible quadratic factors with real coefficients. (b) Factor P completely into linear factors with complex coefficients. \(P(x)=x^{3}-2 x-4\)

Step-by-Step Solution

Answer

(a) \( (x - 2)(x^2 + 2x + 2) \); (b) \( (x - 2)(x + 1 - i)(x + 1 + i) \).

Step 1: Identify Possible Rational Roots

Use the Rational Root Theorem for the polynomial \( P(x) = x^3 - 2x - 4 \). The potential rational roots are factors of the constant term divided by factors of the leading coefficient, which are \( \pm 1, \pm 2, \pm 4 \).

Step 2: Test for Real Roots

Substitute the possible rational roots into the polynomial to check for roots. Trying \( x = 2 \), we find \( P(2) = (2)^3 - 2(2) - 4 = 8 - 4 - 4 = 0 \), so \( x = 2 \) is a root.

Step 3: Perform Synthetic Division

Divide \( P(x) \) by \( x - 2 \) using synthetic division. The quotient is \( x^2 + 2x + 2 \). This is an irreducible quadratic over the reals.

Step 4: Write Real Coefficient Factors

Thus, the factorization with real coefficients is \( P(x) = (x - 2)(x^2 + 2x + 2) \). This satisfies part (a) of the problem.

Step 5: Solve Quadratic for Complex Roots

For part (b), we need the polynomial fully factored into linear factors with complex coefficients. Solve \( x^2 + 2x + 2 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = 2 \).

Step 6: Calculate Complex Roots

Substitute into the quadratic formula: \( x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times 2}}{2 \times 1} = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} = -1 \pm i \).

Step 7: Write Complex Coefficient Factors

Thus, the complete factorization with complex coefficients is \( P(x) = (x - 2)(x + 1 - i)(x + 1 + i) \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a handy tool that helps us in finding the potential rational roots of a polynomial. It states that for a polynomial \[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]with integer coefficients, any rational root, \( \frac{p}{q} \), is such that \( p \) is a factor of the constant term \( a_0 \) and \( q \) is a factor of the leading coefficient \( a_n \). This means to find potential rational roots for \( P(x) = x^3 - 2x - 4 \), we look at the factors of the constant term, \(-4\), and the leading coefficient, \(1\). This gives us \( \pm 1, \pm 2, \pm 4 \) as possible roots. We test these values in the polynomial to identify which ones, if any, provide a true solution, meaning when substituted back, they equate to zero.

Synthetic Division

Synthetic division is a simplified method of dividing a polynomial by a binomial of the form \( x-c \). It's quicker than long division and is especially useful when dealing with polynomials in cases like finding roots.Let's see how it works for our polynomial \( P(x) = x^3 - 2x - 4 \) with a known root, \( x = 2 \). We set this up by writing down the coefficients of \( P(x) \), which are \( 1, 0, -2, -4 \). We then perform the process:

• Place the root \( x = 2 \) to the left of the division.
• Bring down the leading coefficient \( 1 \).
• Multiply it by \( 2 \) and write the result beneath the next coefficient.
• Add these to obtain a sum, continue processing all coefficients.

The final result will show the quotient \( x^2 + 2x + 2 \) as the remaining polynomial when \( x - 2 \) is divided out.

Quadratic Formula

The quadratic formula is a universal method for finding roots of quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our function, the quadratic part \( x^2 + 2x + 2 \) requires finding its roots. Here, \( a = 1 \), \( b = 2 \), and \( c = 2 \).Substituting into the formula, we calculate:\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times 2}}{2 \times 1} \]This simplifies to:\[ x = \frac{-2 \pm \sqrt{-4}}{2} = \frac{-2 \pm 2i}{2} \]Which results in the roots \( x = -1 \pm i \).These roots are complex, which means they involve imaginary numbers where \( i \) is the square root of \(-1\). The use of the quadratic formula provides both solutions systematically without manual trial and error.

Complex Roots

Complex roots occur when a polynomial's discriminant (from the quadratic formula) is negative, resulting in an imaginary number. In our function, the expression \( x^2 + 2x + 2 \) has a discriminant of \(-4\), which leads to complex roots.The roots are \( x = -1 + i \) and \( x = -1 - i \), where \( i \) represents \( \sqrt{-1} \).Every root has a degree of 1 in terms of a linear factor, representing solutions like \( x - (root) \). Therefore, for the polynomial \( P(x) \):

• The linear factor from the real root is \( x - 2 \).
• The complex roots result in two linear factors \( x + 1 - i \) and \( x + 1 + i \).

Putting these together, the complete factorization for both real and complex roots of our polynomial is \( (x - 2)(x + 1 - i)(x + 1 + i) \). This complete factorization shows how complex numbers appear naturally when polynomials are fully broken down.