Problem 56 - Algebra and Trigonometry

Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary, as in Example \(3(\mathrm{a}) .\) $$ P(x)=4 x^{5}-18 x^{4}-6 x^{3}+91 x^{2}-60 x+9 $$

Step-by-Step Solution

Answer

The real zeros of the polynomial are \(x = 3\) and \(x = \frac{1}{2}\).

Step 1: Identify Polynomial Type

The given polynomial is \(P(x) = 4x^5 - 18x^4 - 6x^3 + 91x^2 - 60x + 9\), which is a fifth-degree polynomial. This means there can be up to five real zeros.

Step 2: Use Rational Root Theorem

By the Rational Root Theorem, we can find possible rational zeros by considering factors of the leading coefficient (4) and the constant term (9). The possible rational zeros are \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{9}{4} \).

Step 3: Test Possible Rational Roots

We test each possible rational root by substituting them into the polynomial to see which one results in zero. Upon testing, we find \(x = 3\) and \(x = \frac{1}{2}\) are zeros because they return zero when substituted into the polynomial.

Step 4: Synthetic Division

To simplify the polynomial, perform synthetic division using the discovered zeros \(x - 3\) and \(x - \frac{1}{2}\). First divide the polynomial by \(x - 3\), yielding a new polynomial. Then divide the new polynomial by \(x - \frac{1}{2}\) to reduce it further.

Step 5: Solve Remaining Polynomial

After synthetic division, the remaining polynomial is \(4x^3 - 2x^2 + 3x - 1\). Using the quadratic formula or other factoring techniques may be necessary to solve for the remaining roots. However, for this cubic polynomial, further testing indicates that there are no more rational roots.

Step 6: Calculate Complex Roots

Use numerical or graphical methods, as simply finding rational roots did not yield any more zeros, and factorization is complicated without further constraints or context. The real roots have already been identified by the previous steps.

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a valuable tool when trying to find the real zeros of a polynomial, particularly when dealing with higher-degree polynomials. In simple terms, this theorem helps us identify potential rational roots of a polynomial equation.

• The theorem states that any rational root, expressed as a fraction \( \frac{p}{q} \), must have \(p\) as a factor of the constant term and \(q\) as a factor of the leading coefficient of the polynomial.
• This means that to find possible rational roots, you first list all factors of the constant term (last term) and all factors of the leading coefficient (first term).
• Then form all possible fractions using these factors as numerators and denominators. The resulting list might be long, but it provides a good starting point for finding actual roots.

To illustrate, in our example polynomial \( P(x) = 4x^5 - 18x^4 - 6x^3 + 91x^2 - 60x + 9 \), the constant term is 9 and the leading coefficient is 4. The factors of 9 are \( \pm 1, \pm 3, \pm 9 \), and the factors of 4 are \( \pm 1, \pm 2, \pm 4 \). Combining these gives us the possible rational roots: \( \pm 1, \pm 3, \pm 9, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{9}{2}, \pm \frac{1}{4}, \pm \frac{3}{4}, \pm \frac{9}{4} \).
The theorem doesn't guarantee these numbers are roots, but they are the only candidates for rational roots.

fifth-degree polynomial

A fifth-degree polynomial, as its name suggests, is a polynomial with the highest exponent being five. In other words, it takes the form:
\[ a_{5} x^5 + a_{4} x^4 + a_{3} x^3 + a_{2} x^2 + a_{1} x + a_{0} \] - The degree of the polynomial indicates the maximum number of real roots it can have, so a fifth-degree polynomial can have up to five real roots.
- However, not all these roots must be distinct. There could be repeated roots, or some could be complex, especially since polynomials with real coefficients must have complex roots in conjugate pairs.
- In our example polynomial \( P(x) = 4x^5 - 18x^4 - 6x^3 + 91x^2 - 60x + 9 \), it has already been determined through testing and methods like synthetic division that some zeros are real: specifically, \(x = 3\) and \(x = \frac{1}{2}\).
This is the reasoning behind checking for possible candidates for zeros early in problem-solving, as understanding the nature of your polynomial can guide which methods to use.

synthetic division

Synthetic division is a simplified, quick method for dividing a polynomial by a linear binomial of the form \(x - c\), and it works particularly well when using the zeros discovered via testing.

• Synthetic division involves setting up a simple table of numbers instead of writing out the traditional long division format.
• The process uses the coefficients of the polynomial, simplifying calculations and reducing potential mistakes during division.
• To use this method, write the zero (\(c\)) that corresponds to the factor \((x - c)\) on the left of the table, and the coefficients of the polynomial in a row next to it.
• Then, carry the leading coefficient down, multiply it by the zero, add to the next coefficient, and continue this pattern until you reach the end.

This method can quickly determine the quotient polynomial and verify any remainder. In the example problem, synthetic division was used effectively to test \(x = 3\) and \(x = \frac{1}{2}\) as roots. Removing these roots simplified the problem to solving a cubic, making further analysis or additional techniques easier to apply.

quadratic formula

The quadratic formula is a reliable solution method for finding real roots of a quadratic equation, especially when other methods of factoring or simplifying polynomials do not easily apply.
The formula is given as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Where \( a \), \( b \), and \( c \) are coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). Understanding when to use the quadratic formula is crucial, especially with higher-degree polynomials, because after reducing polynomials through synthetic division and other methods, the remaining parts might be quadratics.

• This formula works by assessing the discriminant \( (b^2 - 4ac) \) to determine the nature of the solutions. If the discriminant is positive, there are two real solutions; if zero, there is one solution; if negative, the solutions are complex.
• Such steps would be necessary after reducing a fifth-degree polynomial, where the remaining polynomial might not factor nicely and direct substitution will reveal no obvious roots.

However, in the exercise above, it was noted through initial testing and synthetic division that the need to use the quadratic formula was bypassed due to the specific characteristics of the polynomial, focusing only on rational roots.