Problem 55 - Precalculus Mathematics for Calculus

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 only real zero of the polynomial is \( x = 1 \).

Step 1: Determine the possible rational roots

According to the Rational Root Theorem, the possible rational roots are the factors of the constant term (12) divided by the factors of the leading coefficient (1). Hence, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).

Step 2: Test possible roots using synthetic substitution

Using synthetic substitution, we test the possible rational roots. We start with \( x = 1 \):\[\begin{array}{r|rrrrrr}1 & 1 & -1 & 7 & -7 & 12 & -12 \ & & 1 & 0 & 7 & 0 & 12 \\hline & 1 & 0 & 7 & 0 & 12 & 0 \\end{array}\]The remainder is 0, so \( x - 1 \) is a factor, and \( x = 1 \) is a root.

Step 3: Perform polynomial division

We can perform polynomial division of \( P(x) \) by \( x - 1 \) to simplify the polynomial:\[P(x) = (x - 1)(x^4 + 7x^2 + 12)\]Now, the task is to find the roots of the quartic polynomial \( x^4 + 7x^2 + 12 \).

Step 4: Substitute and solve by substitution

Let \( y = x^2 \), transforming \( x^4 + 7x^2 + 12 \) into \( y^2 + 7y + 12 = 0 \). Factor this quadratic equation:\( (y + 3)(y + 4) = 0 \).Therefore, the solutions are \( y = -3 \) and \( y = -4 \).

Step 5: Solve for x

Since \( y = x^2 \), substitute back to find \( x \):- For \( y = -3 \), \( x^2 = -3 \), which leads to no real solutions.- For \( y = -4 \), \( x^2 = -4 \), which also leads to no real solutions.Thus, \( x = 1 \) is the only real root.

Key Concepts

Rational Root Theorem

When faced with the task of finding zeros of a polynomial, the Rational Root Theorem is a handy tool. It helps identify possible rational zeros of a polynomial function by considering the factors of the constant term and the leading coefficient. 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 \(1\).

According to the Rational Root Theorem:

• List all factors of the constant term: \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\).
• List all factors of the leading coefficient: \(\pm 1\).

By calculating all possible combinations of factors of the constant term over the factors of the leading coefficient, we identify potential rational roots as \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\). This method reduces the number of candidates you need to test.

Synthetic Substitution

Once you have a set of possible roots from the Rational Root Theorem, testing each quickly is crucial. This can be achieved through synthetic substitution, a streamlined way to evaluate a polynomial at given values.

For instance, to test if \(x = 1\) is a root of \(P(x)\), set up synthetic substitution:

• Write the coefficients of the polynomial: \(1, -1, 7, -7, 12, -12\).
• Use \(x = 1\) as your test value.
• Perform the synthetic substitution process:

The outcome gives \[1 \cdot 1, 0, 7 \cdot 1, 0, 12 \cdot 1, 0\], indicating \(x = 1\) is a valid root since it yields a remainder of zero. This confirms that \(x - 1\) is a factor of the polynomial.

Polynomial Division

Polynomial division simplifies solving polynomials once a root is found. It divides the polynomial by a factor derived from a root, reducing the polynomial's degree.

For example, dividing \(P(x)\) by \(x - 1\) factored as: \[P(x) = (x - 1)(x^4 + 7x^2 + 12)\].

In this way:

• Identify \(x - 1\) as a factor since we've checked that \(x = 1\) is a root.
• Perform the division to simplify \(P(x)\) and obtain a quartic polynomial \(x^4 + 7x^2 + 12\) to solve further.

Polynomial division makes it easier to focus on solving the reduced polynomial, continuing the search for real or complex zeros.

Quartic Polynomial

A quartic polynomial is a fourth-degree polynomial of the form \(ax^4 + bx^3 + cx^2 + dx + e\). In our exercise, after polynomial division, you are left with \(x^4 + 7x^2 + 12\). To solve it, you can make it simpler by a substitution method.

To transform

• Let \(y = x^2\). Now the polynomial becomes \(y^2 + 7y + 12 = 0\).
• Factor this quadratic \((y + 3)(y + 4) = 0\).
• Resolve to get \(y = -3\) and \(y = -4\).

These values signify that the quartic polynomial has no real roots as the squared variables would yield imaginary solutions. However, this transformation step is crucial as it provides a clear path for solving and understanding potential complex solutions.

Roots of a Polynomial

In finding the roots of a polynomial, once you have simplified it, consider the nature of its solutions. After solving our quartic polynomial, we find the solutions \(y = -3\) and \(y = -4\), which imply no real solutions in the context of real numbers.

Translating back, this means solving \(x^2 = -3\) and \(x^2 = -4\) results in no real \(x\) values but complex ones instead.

• No real roots from \(x^2 = -3\) and \(x^2 = -4\).
• Since these equate to imaginary numbers (\(x eq \pm i\sqrt{3}\) and \(x eq \pm i\sqrt{4}\)).

Ultimately, the only real root found is \(x = 1\). Understanding this reinforces that a polynomial's real roots are those values of \(x\) for which \(P(x) = 0\) and visible in real number calculations.