Problem 60 - Algebra and Trigonometry

Question

\(59-64=\) 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 Critical Points for Root Examination

Start by identifying rational roots using the Rational Root Theorem. The theorem states that the possible rational roots of a polynomial are given by the factors of the constant term divided by the factors of the leading coefficient. For \(P(x) = x^3 - 2x - 4\), the constant term is \(-4\) and the leading coefficient is \(1\). Thus, possible rational roots are \(\pm 1, \pm 2, \pm 4\).

Step 2: Check Potential Rational Roots

Evaluate \(P(x)\) for possible rational roots using substitution:- \(P(1) = 1^3 - 2(1) - 4 = -5\)- \(P(-1) = (-1)^3 - 2(-1) - 4 = -1\)- \(P(2) = 2^3 - 2(2) - 4 = 0\)- \(P(-2) = (-2)^3 - 2(-2) - 4 = -4\)From these, \(P(2) = 0\) indicates that \(x = 2\) is a root.

Step 3: Use Synthetic Division

Since \(x = 2\) is a root, perform synthetic division of \(P(x)\) by \(x - 2\) to find the quotient polynomial. Set up the synthetic division:- Coefficients of \(P(x)\) are 1 (for \(x^3\)), 0 (for \(x^2\)), -2 and -4.- Using \(x = 2\):\[\begin{array}{r|rrrr}2 & 1 & 0 & -2 & -4 \ & & 2 & 4 & 4 \\hline & 1 & 2 & 2 & 0 \\end{array}\]The quotient is \(x^2 + 2x + 2\).

Step 4: Factor the Quadratic

We obtained \(x^2 + 2x + 2\) as a quotient, and it cannot be factored further using real coefficients. It can be verified using the discriminant: \(b^2 - 4ac = 2^2 - 4(1)(2) = 4 - 8 = -4\), which indicates complex roots. Thus, it remains as is.

Step 5: Write as Product of Linear and Irreducible Quadratic Factors

Using the results from the steps above, the factorization of \(P(x)\) with real coefficients is \((x - 2)(x^2 + 2x + 2)\). This contains one linear factor and one irreducible quadratic factor.

Step 6: Factor the Quadratic Completely Over Complex Numbers

To factor \(x^2 + 2x + 2\) over the complex numbers, find its roots using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = 2\), and \(c = 2\). Thus:\[x = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i\].This gives the factors \((x + 1 - i)(x + 1 + i)\).

Step 7: Write the Complete Factorization Using Complex Coefficients

Combine the linear factor \(x-2\) with the complex linear factors to write the complete factorization:\[P(x) = (x - 2)(x + 1 - i)(x + 1 + i)\].

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a helpful tool for finding potential rational roots of a polynomial equation. When you have a polynomial like \( P(x) = x^3 - 2x - 4 \), the theorem can guide you to the candidates for rational solutions. The roots are expressed as \( \frac{p}{q} \), where \( p \) divides the constant term and \( q \) divides the leading coefficient.

For \( P(x) \), the constant term is \(-4\), and the leading coefficient is \(1\). Therefore, the possible rational roots could be \( \pm 1, \pm 2, \pm 4 \). Each of these needs to be checked in the polynomial to see if they make \( P(x) = 0 \). In this problem, after testing these values, \( 2 \) is found to be a root because \( P(2) = 0 \). The Rational Root Theorem is an efficient way to start because it reduces the number of values you need to test to a manageable number.

Remember that finding one rational root helps in further polynomial division and simplifies the factorization process.

Synthetic Division

Once you've identified a root, such as \( x = 2 \) for our polynomial \( P(x) \), synthetic division becomes the next powerful tool. This method simplifies polynomial division, especially when dealing with linear roots of the form \( x - c \).

To perform synthetic division on \( P(x) = x^3 - 2x - 4 \), set up with the coefficients \( [1, 0, -2, -4] \) and divide by \( x - 2 \). The synthetic division helps break down the polynomial into a quotient, which further aids in determining the factorization of the polynomial.

• Write the coefficients: \( 1, 0, -2, -4 \).
• Draw a bracket and place the root \( 2 \) on the outside.
• Bring down the first coefficient \( 1 \) to the bottom row.
• Multiply \( 2 \) with the number you just brought down \( 1 \), and write it under the next coefficient.
• Add the column, and repeat the multiplication and addition process.

The result is a simpler quadratic equation \( x^2 + 2x + 2 \), which can be further analyzed or factored with complex roots if necessary.

Quadratic Formula

The Quadratic Formula is essential for finding the roots of any quadratic equation \( ax^2 + bx + c \). When faced with \( x^2 + 2x + 2 \), which we obtained through synthetic division, this formula becomes invaluable. Given

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]
you can calculate the roots of your quadratic. For our equation, \( a = 1 \), \( b = 2 \), and \( c = 2 \). Plug these into the formula to get:
\[ x = \frac{-2 \pm \sqrt{4 - 8}}{2} = \frac{-2 \pm \sqrt{-4}}{2} = -1 \pm i.\]
The resulting roots \(-1 + i\) and \(-1 - i\) indicate that these are complex numbers. Since the discriminant \(b^2 - 4ac\) results in \(-4\), it indicates that the roots are non-real and imaginary.

Complex Numbers

Complex Numbers come into play when the roots of a quadratic equation are not real numbers. In our polynomial \( P(x) = x^3 - 2x - 4 \), after breaking it down into the quadratic part \( x^2 + 2x + 2 \), we apply the Quadratic Formula and find roots that include imaginary components: \(-1 + i\) and \(-1 - i\).

Complex numbers have the form \( a + bi \) where \( i \) is the imaginary unit, satisfying \( i^2 = -1 \). Here, the quadratic portion of the polynomial was transformed into linear complex factors: \((x + 1 - i)(x + 1 + i)\).

• Each root has a real part, \(-1\).
• Each root has an imaginary part, \( \pm i \).

These complex factors efficiently complete the factorization of the original polynomial when combined with the linear factor obtained through rational roots. Understanding these concepts aids in dealing with non-real solutions in polynomial equations.