Problem 50 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The real zeros are 1, -2, \( \frac{-1 + \sqrt{5}}{2} \), and \( \frac{-1 - \sqrt{5}}{2} \).

Step 1: Identify the Polynomial

We are given the polynomial \( P(x) = x^4 + 2x^3 - 2x^2 - 3x + 2 \). Our goal is to find all real zeros, which are values of \( x \) for which \( P(x) = 0 \).

Step 2: Use the Rational Root Theorem

According to the Rational Root Theorem, any rational root of the polynomial is a factor of the constant term divided by a factor of the leading coefficient. Here, the constant term is 2, and the leading coefficient is 1. Therefore, possible rational roots are \( \pm 1, \pm 2 \).

Step 3: Test Possible Rational Roots

We will test the possible roots by substituting into the polynomial:- \( P(1) = 1^4 + 2\times1^3 - 2\times1^2 - 3\times1 + 2 = 1 + 2 - 2 - 3 + 2 = 0 \)- \( P(-1) = (-1)^4 + 2\times(-1)^3 - 2\times(-1)^2 - 3\times(-1) + 2 = 1 - 2 - 2 + 3 + 2 = 2 eq 0 \)- \( P(2) = 2^4 + 2\times2^3 - 2\times2^2 - 3\times2 + 2 = 16 + 16 - 8 - 6 + 2 = 20 eq 0 \)- \( P(-2) = (-2)^4 + 2\times(-2)^3 - 2\times(-2)^2 - 3\times(-2) + 2 = 16 - 16 - 8 + 6 + 2 = 0 \)The roots \( x = 1 \) and \( x = -2 \) are zeros.

Step 4: Divide Polynomial to Find Other Roots

After finding the roots \( x = 1 \) and \( x = -2 \), we will perform polynomial division to factor the polynomial further.First, divide \( P(x) \) by \( x - 1 \). The quotient is \( Q(x) = x^3 + 3x^2 + x - 2 \).Next, divide \( Q(x) = x^3 + 3x^2 + x - 2 \) by \( x + 2 \). The quotient is \( x^2 + x - 1 \).

Step 5: Solve the Quadratic Equation

We need to solve \( x^2 + x - 1 = 0 \) using the quadratic formula.The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 1, c = -1 \).Substitute values:\( x = \frac{-1 \pm \sqrt{1^2 - 4 \times 1 \times (-1)}}{2 \times 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \).The solutions of the quadratic are \( x = \frac{-1 + \sqrt{5}}{2} \) and \( x = \frac{-1 - \sqrt{5}}{2} \).

Step 6: List All Real Zeros

The real zeros of the polynomial \( P(x) \) are \( x = 1, x = -2, x = \frac{-1 + \sqrt{5}}{2} \), and \( x = \frac{-1 - \sqrt{5}}{2} \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a valuable tool when dealing with polynomials. It helps us predict possible rational solutions, which are roots that can be expressed as fractions. To apply the theorem, we take two primary steps:

• First, identify the constant term (the number without a variable) and the leading coefficient (the coefficient of the term with the highest power).
• Next, list all factors of these two numbers. Any potential rational root of the polynomial is a fraction made from these factors, with the constant term on top and the leading coefficient on the bottom.

For example, in the polynomial \( P(x) = x^4 + 2x^3 - 2x^2 - 3x + 2 \), the constant term is 2, and the leading coefficient is 1. Therefore, the possible rational roots are \( \pm 1, \pm 2 \). It’s essential to substitute these values back into the polynomial to test whether they are actual roots. If when substituted, the polynomial equals zero, then it's indeed a root.

Quadratic Formula

The quadratic formula is a standard method for finding the roots of a quadratic equation \( ax^2 + bx + c = 0 \). Sometimes, finding zeros of polynomials requires reducing them down to quadratic form. Here's how the quadratic formula works:

• Identify the coefficients \( a \), \( b \), \( c \) from the quadratic equation.
• Plug these into the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
• The expression under the square root, \( b^2 - 4ac \), is called the discriminant. It determines the nature of the roots:
• If it’s positive, there are two distinct real roots.
• If it's zero, there’s exactly one real root (a double root).
• If negative, the roots are complex numbers, not real numbers.

In the exercise, after performing polynomial division, the polynomial is reduced to \( x^2 + x - 1 = 0 \). Applying the quadratic formula reveals two additional real zeros: \(x = \frac{-1 + \sqrt{5}}{2}\) and \(x = \frac{-1 - \sqrt{5}}{2}\). This demonstrates how effectively the quadratic formula aids in solving quadratics.

Polynomial Division

Polynomial division is akin to long division but involves polynomials. It's a method used to factor polynomials further once some roots are known. The process involves:

• Dividing the original polynomial by a linear factor corresponding to a known root (for example, \( x - r \) where \( r \) is a root).
• Finding the quotient polynomial, which is a degree lower than the original polynomial.
• Repeating the division process if necessary to simplify the polynomial further.

In the given problem, once the roots \( x = 1 \) and \( x = -2 \) are identified using the Rational Root Theorem, we perform polynomial division. First, we divide \( P(x) \) by \( x - 1 \) to get a quotient of \( x^3 + 3x^2 + x - 2 \). Then, we divide the resultant quotient by \( x + 2 \) to obtain \( x^2 + x - 1 \). This process reduces the polynomial to a simpler quadratic form that can be easily solved with the quadratic formula. Polynomial division helps break down complex polynomials into more manageable equations, revealing further roots or confirming existing solutions.