Problem 44 - Algebra 2

Question

CHALLENGE. Consider the polynomial \(f(x)=a x^{4}+b x^{3}+c x^{2}+d x+e,\) where \(a+b+c+d+e=0 .\) Show that this polynomial is divisible by \(x-1\)

Step-by-Step Solution

Answer

The polynomial is divisible by \( x-1 \) since \( f(1) = 0 \).

Step 1: Understanding Polynomial Division by x-1

If a polynomial \( f(x) \) is divisible by \( x-1 \), then \( x=1 \) should be a root of \( f(x) \). This means that when we substitute \( x=1 \) into the polynomial, the result should be 0, i.e., \( f(1) = 0 \).

Step 2: Substitute x=1 into the Polynomial

Substitute \( x=1 \) into the polynomial \( f(x)=a x^{4}+b x^{3}+c x^{2}+d x+e \). This gives us \( f(1) = a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e \). Simplifying this gives \( f(1) = a + b + c + d + e \).

Step 3: Apply Given Condition

We are given that \( a + b + c + d + e = 0 \). Therefore, substituting this into our expression from Step 2, we have \( f(1) = 0 \).

Step 4: Conclude Divisibility

Since \( f(1) = 0 \), \( x=1 \) is a root of the polynomial \( f(x) \). Thus, \( f(x) \) is divisible by \( x-1 \) according to the Factor Theorem.

Key Concepts

Factor Theorem

The Factor Theorem is a useful tool for determining whether a polynomial can be divided by a linear factor without leaving a remainder. Essentially, it connects polynomial roots with factors of a polynomial. This theorem states that if a polynomial \( f(x) \) has a root at \( x = r \), then \( x - r \) is a factor of \( f(x) \). This means you can divide the polynomial by \( x - r \) and get a remainder of zero.

To verify if \( x - 1 \) is a factor of a polynomial, you can use the Factor Theorem by checking if \( x = 1 \) is a root. This involves substituting \( x = 1 \) into the polynomial and checking if the result is zero.

• If \( f(1) = 0 \), then \( x - 1 \) is a factor.
• If \( f(1) eq 0 \), then \( x - 1 \) is not a factor.

By applying the Factor Theorem, we can save time and avoid lengthy polynomial long division when checking for linear factors.

Polynomial Roots

In polynomial mathematics, a root of a polynomial is a value for which the polynomial evaluates to zero. Roots are crucial in the Factor Theorem because they help identify factors of the polynomial. For instance, if \( x = 1 \) is a root of \( f(x) \), it means substituting \( x = 1 \) into \( f(x) \) will result in zero: \( f(1) = 0 \).

Roots can be real or complex numbers depending on the polynomial. For example, the polynomial \( x^2 + 1 \) does not have real roots since it never equals zero for real numbers. Its roots are complex: \( x = i \) and \( x = -i \), where \( i \) is the imaginary unit.

• Real roots result from polynomials that cross or touch the x-axis.
• Complex roots occur in conjugate pairs, such as \( 1 + 2i \) and \( 1 - 2i \).

When dealing with polynomials, finding roots is vital as they suggest the behaviour and characteristics of the polynomial function.

Algebraic Expressions

Algebraic expressions are fundamental elements in mathematics and consist of variables, constants, and arithmetic operations. They are used to construct polynomials, which are specific types of algebraic expressions characterized by terms in the form \( ax^n \). The general form of a polynomial is \( a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \).

Such expressions can represent a variety of mathematical relationships and transformations. For instance, the expression \( ax^4 + bx^3 + cx^2 + dx + e \) is a polynomial with terms of descending powers of \( x \).

• Variables represent unknowns that can take various values.
• Constants are fixed values that do not change.

Working with algebraic expressions—especially polynomials—requires understanding how the coefficients influence the graph of the polynomial. These expressions can be manipulated using arithmetic operations to simplify or factor them, as seen in the original exercise where conditions were used to apply the Factor Theorem efficiently.