Problem 57 - Algebra and Trigonometry

Question

\(57-58\) . Show that the given value(s) of \(c\) are zeros of \(P(x),\) and find all other zeros of \(P(x) .\) $$ P(x)=x^{3}-x^{2}-11 x+15, \quad c=3 $$

Step-by-Step Solution

Answer

The zeros of \(P(x)\) are 3, \(-1 + \sqrt{6}\), and \(-1 - \sqrt{6}\).

Step 1: Verify the Given Value as a Zero

To verify whether \(c = 3\) is a zero of \(P(x)\), substitute \(x = 3\) into \(P(x)\). Calculate \(P(3) = 3^3 - 3^2 - 11 \cdot 3 + 15\). Simplifying this gives \(27 - 9 - 33 + 15 = 0\). Since \(P(3) = 0\), \(c = 3\) is a zero of \(P(x)\).

Step 2: Factor the Polynomial Using the Known Zero

Since \(3\) is a zero, \(x - 3\) is a factor. Use polynomial division to divide \(P(x)\) by \(x - 3\). Performing the division: \((x^3 - x^2 - 11x + 15) \div (x - 3)\) results in a quotient of \(x^2 + 2x - 5\). The factorization is \(P(x) = (x - 3)(x^2 + 2x - 5)\).

Step 3: Solve for Other Zeros of the Quadratic Factor

To find the other zeros, solve \(x^2 + 2x - 5 = 0\) using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), and \(c = -5\). Calculate \(\Delta = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-5) = 4 + 20 = 24\). Thus, \(x = \frac{-2 \pm \sqrt{24}}{2}\). This simplifies to \(x = -1 \pm \sqrt{6}\).

Step 4: List All Zeros of the Polynomial

The zeros of the polynomial \(P(x) = x^3 - x^2 - 11x + 15\) are: \(x = 3\), \(x = -1 + \sqrt{6}\), and \(x = -1 - \sqrt{6}\). These include the given zero and the two additional solutions found from the quadratic factor.

Key Concepts

Polynomial Division

Polynomial division is a useful process when dealing with polynomials, especially when you want to simplify an expression or find its factors. The process is similar to long division with numbers, but instead of digits, you're working with terms that include variables. Here’s how it works:

• Setup: Align the terms of the dividend in descending order of their powers. Place the divisor outside the division symbol.
• Divide the leading term: Divide the first term of the dividend by the first term of the divisor. This gives you the first term of the quotient.
• Multiply and subtract: Multiply the entire divisor by this new quotient term and subtract the result from the dividend. This will bring down a new polynomial.
• Repeat: Repeat the process with the new polynomial formed until the remainder is zero or the degree of the remainder is less than the degree of the divisor.

Once you've gone through these steps, the original polynomial (the dividend) can be expressed as the product of the divisor and the quotient plus the remainder. In our case, dividing \(P(x) = x^3 - x^2 - 11x + 15\) by \(x - 3\) gave us a quotient of \(x^2 + 2x - 5\), showing that \(P(x)\) can be factored as \((x - 3)(x^2 + 2x - 5)\). This factorization is crucial for solving polynomial equations as it helps in finding all the zeros of the polynomial.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). The formula provides a straightforward way to find the values of \(x\) by relying on the coefficients \(a, b,\) and \(c\). The formula itself is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here’s how to use it effectively:

• Calculate the discriminant: The expression under the square root, \(b^2 - 4ac\), is known as the discriminant. It determines the nature of the roots.
  • If the discriminant is positive, there are two distinct real roots.
  • If it's zero, there is one real root (repeated).
  • If it's negative, the roots are complex.
• Plug into the formula: Substitute \(a, b,\) and \(c\) into the quadratic formula.
• Simplify: Simplify the obtained expressions to get the roots.

In our problem, applying the quadratic formula to \(x^2 + 2x - 5 = 0\) resulted in two solutions: \(x = -1 + \sqrt{6}\) and \(x = -1 - \sqrt{6}\). It's an elegant way to find precise solutions when factoring directly isn’t straightforward.

Factoring Polynomials

Factoring polynomials plays a fundamental role in solving polynomial equations. It's about expressing a polynomial as a product of its factors, which are often simpler polynomials. Here's the basic strategy:

• Identify easy factors: If a known zero is given, you can directly obtain a factor, such as \(x - c\) for a zero \(c\).
• Use polynomial division: This is how we find other factors, particularly when you only know one zero or factor.
• Check for patterns: Always be on the lookout for recognizable patterns like difference of squares or perfect square trinomials.

Factoring is particularly important because it simplifies solving. Once a polynomial is expressed in factored form, finding its zeros becomes a task of setting each factor equal to zero and solving. For instance, with the polynomials in our example, we began by factoring \(P(x)\) using a known zero \(x - 3\). Using polynomial division, we further factored it into \((x - 3)(x^2 + 2x - 5)\), allowing us to easily solve for the zeros by finding the roots of each factor.