Problem 31 - Precalculus Mathematics for Calculus

Question

Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. $$P(x)=x^{4}+3 x^{2}-4$$

Step-by-Step Solution

Answer

Zeros are \(x = -1\) and \(x = 1\), each with multiplicity 1.

Step 1: Identify the type of polynomial

The given polynomial is \(P(x) = x^4 + 3x^2 - 4\). Notice that it's a quadratic in form, substituting \(y = x^2\) simplifies it to \(y^2 + 3y - 4\).

Step 2: Factor the quadratic polynomial

We need to factor \(y^2 + 3y - 4\). Look for two numbers that multiply to -4 and add up to 3. These numbers are 4 and -1. Thus, \(y^2 + 3y - 4 = (y + 4)(y - 1)\).

Step 3: Substitute back to original variable

Replace \(y\) with \(x^2\) in the factored form. We get \((x^2 + 4)(x^2 - 1)\). Further factor \(x^2 - 1\) as \((x + 1)(x - 1)\).

Step 4: Calculate the zeros

Set each factor to zero: \(x^2 + 4 = 0\) has no real solutions since it implies \(x^2 = -4\). The factor \((x + 1)(x - 1) = 0\) gives us zeros \(x = -1\) and \(x = 1\).

Step 5: Determine multiplicities

The zeros \(x = -1\) and \(x = 1\) come from \(x^2 - 1\), each has a multiplicity of 1. \(x^2 + 4 = 0\) does not contribute any real zeros.

Key Concepts

Quadratic Polynomial

A quadratic polynomial is an expression of the form \(ax^2 + bx + c\), where \(a, b,\) and \(c\) are constants, and \(a eq 0\). Although it might seem complex, it simply means a polynomial where the highest power of the variable \(x\) is 2.
In the solution, we initially express the degree-4 polynomial \(P(x) = x^4 + 3x^2 - 4\) as a quadratic polynomial by substituting \(y = x^2\). This yields \(y^2 + 3y - 4\), a standard quadratic polynomial.

• Quadratic polynomials have exactly two terms when simplified.
• The name "quadratic" comes from the Latin word for "square," referring to the square of \(x\).
• Factoring quadratic polynomials involves finding two numbers that indicate specific relationships among the terms.

This transformation makes it easier to factor, highlighting the power of substitution to simplify complex expressions.

Zeros of a Polynomial

The zeros of a polynomial are the values of \(x\) that make the polynomial equal to zero. Finding the zeros of a polynomial is like solving the equation \(P(x) = 0\).

In the provided exercise, we aim to find the zeros by setting the fully factored form of \(P(x)= (x^2 + 4)(x + 1)(x - 1)\) to zero. We then solve for \(x\) in each factor:

• For \((x^2 + 4) = 0\), there are no real zeros since \(x^2 = -4\) is not possible with real numbers.
• For \((x + 1) = 0\), solving yields \(x = -1\).
• For \((x - 1) = 0\), solving yields \(x = 1\).

The zeros \(x = -1\) and \(x = 1\) are considered solutions as they satisfy the equation, making the polynomial equal to zero.

Multiplicity of Zeros

The multiplicity of a zero is the number of times it appears as a root of the polynomial. In simpler terms, it shows how many times a factor is repeated in the polynomial's factorization.
In the polynomial \(P(x)\), after factoring, we look at the factored components:

• The factor \((x - 1)\) gives a zero at \(x = 1\) and has a multiplicity of 1 due to its single occurrence.
• Similarly, \((x + 1)\) gives a zero at \(x = -1\), also with a multiplicity of 1.
• \(x^2 + 4 = 0\) has no real zeros, hence no multiplicity for real solutions.

Multiplicities indicate how the graph of the polynomial will behave at each zero.
For example, a zero with an odd multiplicity will cross the x-axis, while an even multiplicity will touch and turn away.