Problem 58
Question
Writing Explain why the terms of \((a-4)^{6}\) have alternating positive and negative signs.
Step-by-Step Solution
Verified Answer
In the expansion of \begin{align*}(a-4)^6\end{align*},the terms have alternating signs because the exponent on \(-4\) alternates between even and odd, affecting the sign due to even powers resulting in a positive sign and odd powers resulting in a negative sign.
1Step 1: Understand the Binomial Theorem
The binomial theorem states that \begin{align*}(x+y)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k\end{align*}where \(\binom{n}{k}\) is the binomial coefficient. The sign of each term in the expansion is determined by the sign of \(x\) and \(y\) being raised to different powers.
2Step 2: Identify the components related to \begin{align*}(a-4)^6\end{align*}
In this case, \(x=a\) and \(y=-4\). Since \(y\) is negative, raising \(-4\) to an even power will result in a positive number, and raising \(-4\) to an odd power will result in a negative number. This alternation of even and odd powers will lead to the alternating positive and negative signs.
3Step 3: Expand the binomial using the components
When \begin{align*}(a-4)^6\end{align*}is expanded using the binomial theorem, the exponents on \(-4\) will alternate between even and odd from one term to the next. Even powers keep the sign of \(-4\) as positive, and odd powers change it to negative, leading to alternating signs in the terms of the expanded binomial.
Key Concepts
Binomial Expansion
Binomial Expansion
At the core of understanding algebraic expressions lies the concept of binomial expansion. This allows us to express the power of a binomial—that is, a two-term algebraic expression— as a sum of terms involving coefficients and powers of the two variables. For example, the binomial expansion of
Other exercises in this chapter
Problem 57
Solve each equation. $$ 2 x^{4}-14 x^{3}+12 x^{2}=0 $$
View solution Problem 57
Find each product. Classify the result by number of terms. $$ (2 c-3)(2 c+4)(2 c-1) $$
View solution Problem 58
How many four-letter permutations can you form from the letters of each word? LINEAR
View solution Problem 58
Solve each equation. $$ 4 x^{4}-2 x^{2}-4=2 $$
View solution