Problem 48 - Precalculus Mathematics for Calculus

Question

Find all zeros of the polynomial. $$P(x)=x^{3}+7 x^{2}+18 x+18$$

Step-by-Step Solution

Answer

The zeros are \( x = -3, -2 + i\sqrt{2}, -2 - i\sqrt{2} \).

Step 1: Identify possible rational roots

To find the zeros of the polynomial, we start by identifying possible rational roots using the Rational Root Theorem. The potential rational roots are the factors of the constant term (18) divided by the leading coefficient (1). These factors are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \). So, the possible rational roots are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).

Step 2: Test possible rational roots

Plug each possible rational root into the polynomial \( P(x) \) to find out if it is a zero. After testing a few possibilities:- For \( x = -3 \), \( P(-3) = (-3)^3 + 7(-3)^2 + 18(-3) + 18 = 0 \). It is a root.By using synthetic division, we can divide the polynomial \( P(x) \) by \( x + 3 \) and find the quotient polynomial.

Step 3: Perform synthetic division

Divide the polynomial by \( x + 3 \) using synthetic division:\(\begin{array}{r|rrr} -3 & 1 & 7 & 18 & 18 \ & & -3 & -12 & -18 \\hline & 1 & 4 & 6 & 0 \\end{array}\)This shows that \( P(x) = (x + 3)(x^2 + 4x + 6) \).

Step 4: Solve the quadratic equation

Now, solve the quadratic \( x^2 + 4x + 6 = 0 \) using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For the quadratic \( a = 1, b = 4, c = 6 \):\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \, \cdot 1 \, \cdot 6}}{2 \, \cdot 1} = \frac{-4 \pm \sqrt{16 - 24}}{2} = \frac{-4 \pm \sqrt{-8}}{2} = \frac{-4 \pm 2i\sqrt{2}}{2} = -2 \pm i\sqrt{2} \]The solutions are \( x = -2 + i\sqrt{2} \) and \( x = -2 - i\sqrt{2} \).

Step 5: List all zeros

Therefore, the zeros of the polynomial \( P(x) = x^3 + 7x^2 + 18x + 18 \) are \( x = -3 \), \( x = -2 + i\sqrt{2} \), and \( x = -2 - i\sqrt{2} \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a helpful tool when you need to find potential rational zeros of a polynomial.
It suggests that any possible rational root of a polynomial equation is a fraction \( \frac{p}{q} \), where:\>

• \( p \) is a factor of the constant term at the end of the polynomial, and
• \( q \) is a factor of the leading coefficient (the coefficient of the term with the highest degree).

This theorem narrows down the number of potential roots to test, which is particularly useful for polynomials with integer coefficients. In our given polynomial \( P(x) = x^3 + 7x^2 + 18x + 18 \), the constant term is 18, and the leading coefficient is 1.
This means our potential rational roots are the factors of 18 divided by 1: \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).
With these possible roots, you can explore which ones actually work in the polynomial equation to find the true rational roots.

synthetic division

Synthetic division offers a simplified method of dividing a polynomial by a linear expression of the form \(x - r\).
It is especially useful for testing possible roots of a polynomial. In our example, after finding \(x = -3\) as a zero using the Rational Root Theorem, we use synthetic division to divide \(P(x)\) by \(x + 3 \) (since \(r = -3\)).

The process is quite straightforward:

• Write down the coefficients of the polynomial: \(1, 7, 18, 18\).
• Use the root \(-3\) and perform the operations along the row of coefficients.
• Drop the first number down as it is, then multiply it by \(-3\), add the result to the next coefficient, and repeat.

The result from synthetic division gives us a quotient, which is a simpler polynomial: \(x^2 + 4x + 6\), and confirms that the division was exact since our remainder is 0.
This polynomial provides the next step in finding other roots.

quadratic formula

Once a polynomial has been reduced using methods like synthetic division, sometimes what's left is a quadratic expression, as seen with \(x^2 + 4x + 6\).
The quadratic formula offers a guaranteed way to find its roots, especially when the roots are not obvious by factoring. The formula is expressed as:

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

In this instance, we have \(a = 1\), \(b = 4\), and \(c = 6\).

• Substitute these values into the formula to compute:
• \(x = \frac{-4 \pm \sqrt{16 - 24}}{2} = \frac{-4 \pm \sqrt{-8}}{2}\)

This yields complex roots due to the negative discriminant (\(-8\)).
The results are \(x = -2 + i\sqrt{2}\) and \(x = -2 - i\sqrt{2}\), showing that complex roots can also emerge from real polynomials. Understanding and using the quadratic formula gives us the complete set of solutions for our polynomial, combining real and complex numbers.