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Chapter 12: Problem 43

\(43-44\) Simplify using the Binomial Theorem. $$ \frac{(x+h)^{3}-x^{3}}{h} $$

Short Answer

Expert verified
The simplified expression is \\( 3x^2 + 3xh + h^2 \\\).

Step by step solution

01

Recall the Binomial Theorem

The Binomial Theorem states \( (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k \). In this exercise, apply the binomial theorem to \( (x + h)^3 \).
02

Apply the Binomial Expansion

Expand \( (x + h)^3 \) using the Binomial theorem: \( (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \).
03

Substitute and Simplify

Substitute \( (x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \) into the original expression \( \frac{(x+h)^{3}-x^{3}}{h} \). This gives \( \frac{x^3 + 3x^2h + 3xh^2 + h^3 - x^3}{h} \).
04

Cancel Similar Terms

Simplify the expression by canceling \( x^3 - x^3 \), resulting in \( \frac{3x^2h + 3xh^2 + h^3}{h} \).
05

Factor and Simplify

Factor out \( h \) from the numerator: \( \frac{h(3x^2 + 3xh + h^2)}{h} \). Cancel \( h \) from the numerator and denominator, which simplifies to \( 3x^2 + 3xh + h^2 \).

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Key Concepts

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

Polynomial Expansion
The polynomial expansion of a binomial expression involves expressing the expression as a sum of terms. This is achieved using the Binomial Theorem, which provides a formula for expanding a binomial raised to any power. The theorem can be written as \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\). This allows us to expand expressions like \((x + h)^3\) in a systematic way.
  • The terms generated are based on the powers of each variable, following the order from the highest power to the lowest.
  • Each term is made up of a combination of coefficients given by the binomial coefficients \(\binom{n}{k}\), which can be found in Pascal's triangle.
The polynomial \((x + h)^3\) expands to \(x^3 + 3x^2h + 3xh^2 + h^3\), capturing the interaction of \(x\) and \(h\) in powers that sum to 3. Understanding polynomial expansion is crucial for simplifying expressions efficiently, especially when dealing with higher powers.
Simplification Techniques
Simplifying expressions means reducing them to their simplest or most basic form without altering their value. In many exercises, especially using the Binomial Theorem, simplification involves several key strategies:
  • Cancellation: Once expansions occur, identify terms that cancel each other out. This helps reduce the complexity of the expression. For instance, in \(\frac{(x+h)^3-x^3}{h}\), after expanding and simplifying, \(x^3\) terms cancel out.
  • Factorization: Look for common factors in the terms. Factorization allows you to simplify through division. In this problem, terms were factored by \(h\), allowing for cancellation in the fraction.
  • Reduction of Fractions: When numerators and denominators share factors, cancel them to simplify the fraction. This is the final step after expansion in the original exercise.
Simplification not only makes expressions easier to understand but also prepares them for further mathematical operations. In this case, the expression is narrowed down to \(3x^2 + 3xh + h^2\), making it more manageable for analysis or further calculations.
Algebraic Expressions
Algebraic expressions are combinations of variables, numbers, and operations. They form the basis of algebra and are fractionally linked to the concepts of polynomial expansion and simplification. In exercises involving expressions like \((x + h)^3\), you work with an expression that transforms through various algebraic processes:
  • Terms and Coefficients: Understanding how terms (with variables raised to exponents) and coefficients (numbers that multiply the terms) interact is key. Each term in \((x+h)^3\) contributes to the overall expression.
  • Operations: Operations like addition, subtraction, multiplication, and division are used to manipulate the terms. Identifying the sequence of operations helps maintain the integrity of the expression.
  • Structure: Well-structured expressions adhere to algebraic rules, allowing for predictable outcomes when simplifying. This structural understanding assists in reducing complex expressions to simple forms.
In exercises like the given one, practicing the handling of algebraic expressions aids in recognizing patterns, simplifies processing, and prepares you for solving more complex algebraic problems. Mastery of these expressions enables students to engage confidently in algebraic manipulation across various contexts.

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Most popular questions from this chapter

\(1-12\) . Use Pascal's triangle to expand the expression. $$ \left(x^{2} y-1\right)^{5} $$

Express the repeating decimal as a fraction. $$ 0.2 \overline{53} $$

\(13-20=\) Evaluate the expression. $$ \left(\begin{array}{c}{100} \\ {98}\end{array}\right) $$

Amortizing a Mortgage When they bought their house, John and Mary took out a \(\$ 90,000\) mortgage at 9\(\%\) interest, repayable monthly over 30 years. Their payment is \(\$ 724.17\) per month (check this using the formula in the text). The bank gave them an amortization schedule, which is a table showing how much of each payment is interest, how much goes toward the principal, and the remaining principal after each payment. The table below shows the first few entries in the amortization schedule. $$ \begin{array}{|c|c|c|c|c|}\hline \text { Payment } & {\text { Total }} & {\text { Interest }} & {\text { Principal }} & {\text { Remaining }} \\\ {\text { number }} & {\text { payment }} & {\text { payment }} & {\text { payment }} & {\text { principal }} \\ \hline 1 & {724.17} & {675.00} & {49.54} & {89,950.83} \\ {2} & {724.17} & {674.63} & {49.54} & {89,901.29} \\\ {3} & {724.17} & {674.26} & {49.91} & {89,851.38} \\ {4} & {724.17} & {673.89} & {50.28} & {89,801.10} \\ \hline\end{array} $$ After 10 years they have made 120 payments and are wondering how much they still owe, but they have lost the amortization schedule. (a) How much do John and Mary still owe on their mortgage? [Hint: The remaining balance is the present value of the 240 remaining payments. (b) How much of their next payment is interest and how much goes toward the principal? [Hint: Since 9\(\% \div\) \(12=0.75 \%\) , they must pay 0.75\(\%\) of the remaining principal in interest each month.

Annuity Find the amount of an annuity that consists of 20 semiannual payments of \(\$ 500\) each into an account that pays 6\(\%\) interest per year, compounded semiannually.
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