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Chapter 12: Problem 40

Without expanding, what is the leading term of $$ (2 s+5)(3 s+1)(s-10) ? $$

Short Answer

Expert verified
Answer: The leading term of the polynomial is \(6s^3\).

Step by step solution

01

Identify the leading terms of each binomial

In each binomial, the leading term is the term with the highest power of the variable \(s\): - For the first binomial \((2s+5)\), the leading term is \(2s\) - For the second binomial \((3s+1)\), the leading term is \(3s\) - For the third binomial \((s-10)\), the leading term is \(s\)
02

Multiply the leading terms

Now, to get the leading term of the entire polynomial, multiply the leading terms we identified in step 1: $$(2s) \times (3s) \times (s)$$
03

Simplify the leading term

Finally, simplify the expression we got in step 2: $$6s^3$$ Thus, the leading term of the given polynomial is \(\boxed{6s^3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Binomial Leading Term
When dealing with binomials, it's important to identify the leading term. The leading term is the term with the highest power of the variable.
In binomial expressions like \((2s + 5)\), \((3s + 1)\), and \((s - 10)\), you look for the term involving the variable raised to the highest power, which is usually the first term.

For example:
  • In \((2s + 5)\), the leading term is \(2s\), because the "\(s\)" is of the first power, and it is the main focus compared to the constant \(5\).
  • In \((3s + 1)\), the leading term is \(3s\).
  • In \((s - 10)\), the leading term is \(s\).

The leading terms form the building blocks to determine the overall behavior of the polynomial when multiplied with each other.
Multiplying Binomials
To understand how binomials interact, we often multiply them. This process involves dealing with each part of the expression, especially their leading terms.

Suppose you want to find the leading term of the product of several binomials like in our example:
  • The leading term of \((2s + 5)(3s + 1)(s - 10)\) can be found by multiplying the leading terms: \((2s)\), \((3s)\), and \((s)\).
Here's how multiplication of the leading terms works:
\[ (2s) \times (3s) \times (s) = 6s^3 \]
This focuses on multiplying the coefficients \(2\), \(3\), and the implicit \(1\) in \(s\) together, and then combining the powers of \(s\).

The result \(6s^3\) represents the highest degree term of the new polynomial created by multiplying the three binomials together.
Simplifying Expressions
Simplifying polynomial expressions involves reducing them to their most basic form. This often means combining like terms and eliminating redundancies.

In the context of leading terms, the goal is to ensure you're dealing with the most straightforward representation of the term. After multiplying the leading terms,
  • We find \(6s^3\) as the simplified leading term.
This result comes from:
  • Combining all the coefficients: \((2 \times 3 = 6)\).
  • Adding the powers of \(s\): \(1 + 1 + 1 = 3\), giving us \(s^3\).

The simplified form \(6s^3\) is crucial because it tells us how this polynomial behaves for large values of \(s\). It reflects both the degree and the leading coefficient, providing insight into the polynomial's growth and direction.

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Most popular questions from this chapter

Give all the solutions of the equations. $$ (s+10)^{2}-6(s+10)-16=0 $$

Refer to Example 2 on page 379 about the value of annual gifts to Elliot growing at an annual growth factor of \(x=1+r,\) where \(r\) is the annual interest rate. The total value of his investments on his \(20^{\text {th }}\) birthday is $$1000 x^{5}+500 x^{4}+750 x^{3}+1200 x+650$$ (a) What were the gifts on his \(18^{\text {th }}, 19^{\text {th }}\) and \(20^{\text {th }}\) birthdays? (b) Evaluate the polynomial in part (a) for \(x=\) \(1.05,1.06,1.07 .\) What do these values tell you about the investment?

Refer to the functions \(f(x)\) and \(g(x),\) where the function $$ g(x)=1+\frac{1}{2} x+\frac{3}{8} x^{2}+\frac{5}{16} x^{3} $$ is used to approximate the values of $$ f(x)=\frac{1}{\sqrt{1-x}} $$ Show that \(f(x)\) is undefined at \(x=1\) and \(x=2\), but that \(g(x)\) is defined at these values. Explain why the algebraic operations used to define \(f\) may lead to undefined values, whereas the operations used to define \(g\) will not.

List the nonzero coefficients of the polynomials. $$ \frac{s^{13}}{3} $$

Find the constants \(r, s, p,\) and \(q\) if multiplying out the polynomial \(\left(r x^{5}+2 x^{4}+3\right)\left(2 x^{3}-s x^{2}+p\right)\) gives \(6 x^{8}-11 x^{7}-10 x^{6}-12 x^{5}-8 x^{4}+q x^{3}-15 x^{2}-12 .\)
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