Problem 55 - Algebra and Trigonometry

Question

Find all zeros of the polynomial. \(P(x)=x^{5}-x^{4}+7 x^{3}-7 x^{2}+12 x-12\)

Step-by-Step Solution

Answer

The zeros are 1, \(\pm i\sqrt{3}\), \(\pm 2i\).

Step 1: Identify Possible Rational Roots

We start by using the Rational Root Theorem, which suggests that any rational solution of the polynomial equation \(P(x)=0\) is a factor of the constant term (12) divided by a factor of the leading coefficient (1). Thus, the potential rational roots are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\).

Step 2: Test Possible Rational Roots Using Synthetic Division

Test these possible roots by substituting them into the polynomial or using synthetic division. Upon testing, we find that \(x=1\) is a root because substituting \(x=1\) into the polynomial gives zero or the remainder from synthetic division is zero. Therefore, \(x-1\) is a factor of \(P(x)\).

Step 3: Perform Polynomial Division to Find Quotient

Use synthetic division to divide the polynomial \(P(x)\) by \(x-1\). This gives us a quotient polynomial \(Q(x)=x^4+7x^3+12x^2+12\).

Step 4: Factor or Solve the Quotient Polynomial

Attempt to factor the quotient polynomial \(Q(x)\). This leads to the finding of another factor \((x^2 + 3)(x^2 + 4)\). These quadratic expressions can be potentially solved using the quadratic formula. For \(x^2 + 3 = 0\), solutions are \(x = \pm \sqrt{-3} = \pm i\sqrt{3}\). For \(x^2 + 4 = 0\), solutions are \(x = \pm \sqrt{-4} = \pm 2i\).

Step 5: Compile the Solutions

The complete set of zeros of the polynomial \(P(x)\) therefore includes both the real and complex roots derived from previous steps: \(x=1, x=\pm i\sqrt{3}, x=\pm 2i\).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a helpful tool in determining the potential rational zeros of a polynomial. This theorem tells us that any rational solution of a polynomial equation is a fraction whose numerator is a factor of the polynomial's constant term and whose denominator is a factor of the leading coefficient.
This allows for a finite number of possibilities to check, making the task of finding rational roots more manageable.

For example, in the polynomial \(P(x) = x^5 - x^4 + 7x^3 - 7x^2 + 12x - 12\), the constant term is 12, and the leading coefficient is 1. This means that the potential rational roots are \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\). Simply put, we can determine these as follows:

• Factors of the constant term (12): \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\)
• Factors of the leading coefficient (1): \(\pm 1\)

Combining these, we get the list of possible rational roots. However, it doesn't guarantee that all of them are actual roots, only that these are the possibilities to test.

synthetic division

Synthetic division is a streamlined version of polynomial division, specially used when dividing by a linear factor of the form \(x - c\). It involves a sequence of operations in a tabular form, which is simpler and quicker than the standard polynomial division method.

To perform synthetic division, follow these steps:

• Write the coefficients of the polynomial in descending order, noting any missing terms with a \(0\).
• Use the potential root (from Rational Root Theorem) as the divisor.
• Bring down the leading coefficient as it is.
• Multiply the divisor by the number just written below the line, and carry it to the next column to add it to the coefficient present above.
• Repeat this process for each column.

If the final result (remainder) is zero, the divisor is indeed a root of the polynomial. For instance, when we test \(x = 1\) as a root for the polynomial \(P(x)\), performing synthetic division showed us that the remainder is zero, confirming \(x = 1\) is a valid root. The division also provided the quotient polynomial, which is then used for further factorization.

quadratic formula

The Quadratic Formula is a powerful tool used to find solutions to quadratic equations of the form \(ax^2 + bx + c = 0\). It is especially useful when the quadratic does not factor neatly. The formula given by:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

allows us to determine the roots of any quadratic equation effectively, including those resulting in complex numbers.

In the process of solving the polynomial \(P(x)\), after we found that \(x = 1\) was a root, we were left with the quartic polynomial \(Q(x) = x^4 + 7x^3 + 12x^2 + 12\). This was further factorized into two quadratics: \(x^2 + 3\) and \(x^2 + 4\). These are solvable using the quadratic formula, showing:

• For \(x^2 + 3 = 0\), we solve: \(x = \pm \sqrt{-3} = \pm i\sqrt{3}\)
• For \(x^2 + 4 = 0\), we solve: \(x = \pm \sqrt{-4} = \pm 2i\)

These results illustrate how we can find both real and complex roots, using the quadratic formula on irreducible quadratics. Each root provides a necessary component in understanding the full nature of the polynomial's zeroes.