Problem 44 - Algebra and Trigonometry

Question

\(41-58=\) Find all zeros of the polynomial. $$ P(x)=x^{3}+7 x^{2}+18 x+18 $$

Step-by-Step Solution

Answer

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

Step 1: Overview

To find all zeros of the polynomial \(P(x)=x^{3}+7 x^{2}+18 x+18\), we need to determine the values of \(x\) for which \(P(x) = 0\). These values are known as the roots or zeros of the polynomial.

Step 2: Check for Rational Roots

According to the Rational Root Theorem, any rational root, \(\frac{p}{q}\), of the polynomial equation must be a factor of the constant term \(18\) (numerator \(p\)) divided by a factor of the leading coefficient, which is \(1\) (denominator \(q\)). The possible rational roots are thus \(\pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18\).

Step 3: Test Possible Roots Using Synthetic Division

Evaluate each possible rational root using synthetic division or direct substitution to check if they are actual roots. For example, testing \(x = -1\), the calculation gives:\[P(-1) = (-1)^3 + 7(-1)^2 + 18(-1) + 18 = -1 + 7 - 18 + 18 = 6, \]which is not zero. Repeat this process for other possible roots.

Step 4: Identify Actual Root

Upon testing possible roots, you may find \(x = -3\) as an actual root because \[P(-3) = (-3)^3 + 7(-3)^2 + 18(-3) + 18 = -27 + 63 - 54 + 18 = 0.\]Thus, \(x = -3\) is a root of the polynomial.

Step 5: Factor the Polynomial

Since \(x = -3\) is a root, divide the polynomial by \(x + 3\) using synthetic or long division to find the other factors. After performing synthetic division, the quotient is \(x^2 + 4x + 6\), which means \(P(x) = (x + 3)(x^2 + 4x + 6)\).

Step 6: Solve the Quadratic Equation

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

Step 7: Conclusion

The zeros of the polynomial \(P(x)\) are \(x = -3,\ x = -2 + i\sqrt{2},\) and \(x = -2 - i\sqrt{2}\). These are all the values of \(x\) for which \(P(x) = 0\).

Key Concepts

Rational Root Theorem

When trying to find the zeros of a polynomial, the Rational Root Theorem can be a very handy tool. This theorem helps us predict possible rational zeros (or roots) of a polynomial function. The idea behind the Rational Root Theorem is quite simple: if a polynomial has any rational roots, those roots must be of the form \( \frac{p}{q} \), where \( p \) is a factor of the constant term (the term without \( x \)) and \( q \) is a factor of the leading coefficient (the coefficient of the term with the highest power of \( x \)). For a polynomial like \( P(x) = x^3 + 7x^2 + 18x + 18 \):

• The constant term is 18, and its factors are \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \).
• The leading coefficient of \( x^3 \) is 1, and its factors are just \( \pm 1 \).

Therefore, the possible rational roots of this polynomial are the factors of 18 divided by the factors of 1, which still leaves us with \( \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18 \). This provides us with a list of numbers to check through substitution or by using synthetic division.

Synthetic Division

Once we have a list of potential rational zeros, synthetic division can help verify if any of them truly are roots without going through more complicated calculations. Synthetic division is a simplified form of polynomial long division and is more efficient when dealing with linear divisors.Imagine using synthetic division to test \( x = -3 \) as a root for the polynomial \( P(x) = x^3 + 7x^2 + 18x + 18 \). Here’s how it works:

• Write down the coefficients of the polynomial: \( 1, 7, 18, 18 \).
• Place \( -3 \) (our test value) to the left of the coefficients and start the synthetic division process.
• Bring down the first coefficient (1) as it is.
• Multiply this number by \( -3 \), place the result below the next coefficient, and add down.
• Repeat this multiply-add process through all coefficients.

If the final result (called the remainder) is 0, then \( x + 3 \) is a factor, confirming \( x = -3 \) as a root. For \( P(x) \), this process shows zero remainder, proving \( x = -3 \) is indeed a root and providing a quotient polynomial, \( x^2 + 4x + 6 \), for further analysis.

Quadratic Formula

After using synthetic division, you might end up with a simpler polynomial, like a quadratic, that needs further attention. If a quadratic equation remains, solving it is essential to find the remaining zeros of the original polynomial. For a quadratic equation like \( x^2 + 4x + 6 = 0 \), the Quadratic Formula comes into play:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this equation, \( a \), \( b \), and \( c \) are the coefficients from the equation \( ax^2 + bx + c = 0 \).

For our case here, \( a = 1 \), \( b = 4 \), and \( c = 6 \).

Plug these into the formula:

• Calculate the discriminant: \( b^2 - 4ac = 4^2 - 4 \times 1 \times 6 = 16 - 24 = -8 \).
• Since the discriminant is negative, the solutions include imaginary numbers: \(-2 \pm i\sqrt{2} \).

Thus, the equation yields two complex roots: \( -2 + i\sqrt{2} \) and \( -2 - i\sqrt{2} \), completing the set of zeros for the original polynomial together with the rational zero found earlier, \( x = -3 \). It shows how vital the Quadratic Formula is when finding both real and complex roots.