Problem 46 - Precalculus Mathematics for Calculus

Question

Find all zeros of the polynomial. $$P(x)=x^{3}-7 x^{2}+17 x-15$$

Step-by-Step Solution

Answer

The zeros are \( x = 3 \), \( x = 2+i \), and \( x = 2-i \).

Step 1: Identify the Polynomial

We need to find the zeros of the polynomial \( P(x) = x^3 - 7x^2 + 17x - 15 \). A zero of the polynomial is any value \( x \) such that \( P(x) = 0 \).

Step 2: Use the Rational Root Theorem

Apply the Rational Root Theorem, which says if \( P(x) \) has a rational zero \( \frac{p}{q} \), then \( p \) is a factor of the constant term (-15) and \( q \) is a factor of the leading coefficient (1). Thus, possible rational zeros are \( \pm 1, \pm 3, \pm 5, \pm 15 \).

Step 3: Test Potential Zeros

Evaluate \( P(x) \) for the potential zeros by substitution: - \( P(1) = 1^3 - 7(1)^2 + 17(1) - 15 = -4 \).- \( P(3) = 3^3 - 7(3)^2 + 17(3) - 15 = 0 \).3 is a zero.

Step 4: Factor the Polynomial Using Synthetic Division

Now that we know \( x = 3 \) is a zero, use synthetic division to divide \( P(x) \) by \( x-3 \).Perform synthetic division:| 3 | 1 | -7 | 17 | -15 ||--|----|----|----|-----|| | | 3 | -12 | 15 ||--|----|----|----|-----|| 1 | -4 | 5 | 0 |This gives us the quotient \( x^2 - 4x + 5 \) and a remainder of 0.

Step 5: Solve the Quadratic Equation

We have reduced the problem to solving \( x^2 - 4x + 5 = 0 \). Use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \(a = 1, b = -4, c = 5\):\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} \]\[ x = \frac{4 \pm \sqrt{16 - 20}}{2} \]\[ x = \frac{4 \pm \sqrt{-4}}{2} \]\[ x = \frac{4 \pm 2i}{2} \]\[ x = 2 \pm i \]

Step 6: Compile the Zeros

The zeros of the polynomial \( P(x) \) are \( x = 3 \), \( x = 2+i \), and \( x = 2-i \). These represent all the solutions to the equation \( P(x) = 0 \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool that helps us find possible rational zeros of a polynomial equation. It provides us a way to narrow down potential solutions into a list of rational numbers. This is especially useful when we're dealing with polynomials with whole number coefficients.

The theorem states that if a polynomial has a rational zero of the form \( \frac{p}{q} \), then \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient. In the exercise, the polynomial \( P(x) = x^3 - 7x^2 + 17x - 15 \) has a constant term of \(-15\) and a leading coefficient of \(1\).

• Possible values for \( p \) given that \( p \) is a factor of \(-15\) include: \( \pm 1, \pm 3, \pm 5, \pm 15 \).
• The only factor of the leading coefficient \( q = 1 \) is \( \pm 1 \).

Thus, possible rational roots for the polynomial are simply those numbers that are also factors of the constant term: \( \pm 1, \pm 3, \pm 5, \pm 15 \). By checking these candidates, we shortened our list of possible solutions to check more easily.

Synthetic Division

Synthetic division is a simplified form of polynomial division, especially when dividing by a linear factor. It is much quicker than long division and perfect for scenarios where you're testing possible roots, found using the Rational Root Theorem.

When applying synthetic division, you are essentially evaluating the polynomial at the root and breaking it down to find the quotient and remainder.

For instance, in this exercise, we tested \( x = 3 \) and found it to be a root because when substituting into the polynomial, it results in zero. We then use synthetic division to divide \( P(x) = x^3 - 7x^2 + 17x - 15 \) by \( x - 3 \).

Here's a brief overview of the steps:

• Write down the coefficients of the polynomial: \( 1, -7, 17, -15 \).
• Place \( 3 \) (the root) to the left.
• Bring down the leading coefficient as it is.
• Multiply it by 3, the root, and add it to the next coefficient.
• Continue this pattern until you finish.

The result we obtained was a quotient of \( x^2 - 4x + 5 \) and a remainder of zero, confirming \( x - 3 \) is a factor of the polynomial.

Quadratic Formula

To find the zeros of a quadratic equation, we commonly use the quadratic formula. This formula helps solve any quadratic equation in the standard format: \( ax^2 + bx + c = 0 \). In our case, after synthetic division, we found ourselves with the quadratic \( x^2 - 4x + 5 = 0 \).

The quadratic formula is:

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

• \( a = 1 \)
• \( b = -4 \)
• \( c = 5 \)

We substitute these values into the formula:
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(5)}}{2(1)} \]
After computing, the solutions are \( x = 2 \pm i \), indicating complex roots, with \( i \) being the imaginary unit \( \sqrt{-1} \). This means the polynomial is non-factorable over the real numbers, and the roots reflect the nature of complex numbers. It also emphasizes that not all zeros of polynomials are rational or even real.