Problem 32 - College Algebra and Calculus: An Applied Approach

Question

Find all the zeros of the function and write the polynomial as a product of linear factors. $$f(x)=x^{4}+29 x^{2}+100$$

Step-by-Step Solution

Answer

The zeros of function \(f(x)=x^{4}+29 x^{2}+100\) are \(\pm 3, \pm \sqrt{11}i\). The function written as a product of linear factors is \(f(x) = (x-3)(x+3)(x-\sqrt{11}i)(x+\sqrt{11}i)\).

Step 1: Simplify the Polynomial

The given polynomial function can be written as \(f(x)=x^{4}+29 x^{2}+100\). If we let \(z = x^{2}\) then the function takes shape of a quadratic equation, and transforms into \(f(z)=z^{2}+29z+100\).

Step 2: Apply the Quadratic Formula

Apply the Quadratic formula to find the roots of the simplified equation. The quadratic formula is given by: \[ z = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \], where in this case \(a = 1\), \(b = 29\), and \(c = 100\). Substituting these values into the formula, we get: \[ z = \frac{-29 \pm \sqrt{(29)^{2} - 4 \cdot 1 \cdot 100 }}{2 \cdot 1}, \] \[ z = \frac{-29 \pm \sqrt{641}}{2}, \] So, the roots of the equation are: \( z_{1} = 9, z_{2} = -11 \).

Step 3: Substitute z with \(x^{2}\)

Remember that we transformed the polynomial function into a quadratic function by letting \(z = x^{2}\). Now it's time to revert this substitution. Since \(z_{1} = 9 = x^{2}_{1}\), we can imply that the roots are \(x_{1} = \sqrt{9} = \pm 3\). Similarly, as \(z_{2} = -11 = x^{2}_{2}\), the roots are \(x_{2} = \sqrt{-11} = \pm \sqrt{11}i \) where \(i=\sqrt{-1}\).

Step 4: Presentation of the function as a product of its zeros

Express the original polynomial as a product of linear factors using the roots obtained. So, \(f(x)=x^{4}+29 x^{2}+100 = (x-3)(x+3)(x-\sqrt{11}i)(x+\sqrt{11}i)\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for finding the roots of quadratic equations. A quadratic equation is one that can be written in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides a straightforward method to compute the roots, which could be real or complex. The expression under the square root, \( b^2 - 4ac \), is known as the discriminant. It determines the nature of the roots:

• If it is positive, there are two distinct real roots.
• If it is zero, there is one real root, which means the roots are repeated.
• If it is negative, the roots are complex.

In the example provided, this formula was used after the original polynomial was rewritten into a quadratic equation. Substituting \( a = 1 \), \( b = 29 \), and \( c = 100 \), we calculated the roots of the equation \( z^2 + 29z + 100 = 0 \). This ultimately helped us in finding the zeros of the original polynomial after substituting back \( z = x^2 \).

Linear Factors

A polynomial can be expressed as a product of linear factors once we know its roots. Linear factors are expressions of the form \( x - r \), where \( r \) is a root of the polynomial. If a polynomial has a degree \( n \), it can be written as a product of \( n \) linear factors.To write a polynomial as a product of its linear factors:

• Find all the roots of the polynomial.
• Write each root as a linear factor \( (x - r) \).

In the given exercise, the roots of the polynomial \( f(x) = x^4 + 29x^2 + 100 \) were found to be \( x = 3, -3, \sqrt{11}i, -\sqrt{11}i \). Hence, the polynomial can be expressed as a product of these linear factors:

• \( (x-3)(x+3)(x-\sqrt{11}i)(x+\sqrt{11}i) \)

This factorization is helpful not only for solving equations but also for understanding the behavior of the polynomial function.

Complex Roots

Complex roots arise when the discriminant \( b^2 - 4ac \) is negative in the quadratic formula. A complex number includes a real part and an imaginary part, denoted by \( i \), which is the square root of \( -1 \). For two complex roots, they always occur in conjugate pairs. This means if \( a + bi \) is a root, \( a - bi \) will also be a root.In the exercise, the root \( z_2 = -11 \) led to the complex roots \( x_{2} = \pm \sqrt{11}i \), which are \( \sqrt{11}i \) and \( -\sqrt{11}i \). Here:

• \( \sqrt{-11} = \sqrt{11}i \)
• \( i^2 = -1 \) which neatly resolves the negative under the square root.

These roots were then used to form part of the polynomial's linear factorization. Complex roots play a vital role in expanding polynomials into products of linear factors, especially for higher degree polynomials where not all roots are real.