Problem 68 - Algebra Form and Function

Question

Give the value of \(a\) that makes the statement true. The degree of \((t-1)^{3}+a(t+1)^{3}\) is less than 3

Step-by-Step Solution

Answer

Question: Determine the value of 'a' that makes the degree of the polynomial \((t-1)^{3}+a(t+1)^{3}\) less than 3. Answer: The value of 'a' that makes the degree of the polynomial less than 3 is -1.

Step 1: Expanding the expression

Expand the polynomial \((t-1)^{3}+a(t+1)^{3}\) using the binomial theorem or by simply multiplying the terms. \((t-1)^{3} = t^3 - 3t^2 + 3t - 1\) \(a(t+1)^{3} = at^3 + 3at^2 + 3at + a\) The expanded expression becomes: \((t^3 - 3t^2 + 3t - 1) + (at^3 + 3at^2 + 3at + a)\)

Step 2: Combining like terms

Group the terms with the same degree together to simplify the expression. \((t^3 + at^3) + (-3t^2 + 3at^2) + (3t + 3at) + (-1 + a)\)

Step 3: Coefficient of the highest-degree term

For the degree of the polynomial to be less than 3, the coefficient of the \(t^3\) term must be zero, meaning that the combination of the coefficients of \(t^3\) should be zero. So, we have: \(1+a = 0\)

Step 4: Solve for 'a'

Solve the equation \(1+a=0\) to find the required value of 'a'. \(a = -1\) The value of \(a\) that makes the degree of the polynomial \((t-1)^{3}+a(t+1)^{3}\) less than 3 is \(\boxed{-1}\).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful algebraic tool that helps in expanding expressions raised to a power. It is particularly useful for expressions of the form \((a + b)^n\). The theorem states that:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]Here, \(\binom{n}{k}\) is a binomial coefficient, which can be calculated using combinations. In simpler terms, the theorem provides a formula to expand any power of a binomial into a sum involving terms of the original binomial with varying coefficients.

• Each term in the expansion has the format: \(\binom{n}{k} a^{n-k} b^k\)
• In our example, for \((t-1)^3\) and \((t+1)^3\), the binomial theorem helps us derive each term of the expanded polynomial effortlessly.

By substituting into the formula and simplifying, we obtained the expanded forms:- \((t-1)^3 = t^3 - 3t^2 + 3t - 1\) - \(a(t+1)^3 = at^3 + 3at^2 + 3at + a\) Breaking down expressions using the binomial theorem simplifies complex calculations and makes polynomial manipulations more manageable.

Combining Like Terms

Combining like terms is a crucial step in simplifying polynomial expressions. After expanding each part of a polynomial, you'll often end up with several terms that have the same degree. By grouping these terms, we can simplify the expression, showing only one term per degree.

• "Like terms" are terms that have the same variable raised to the same power.
• In our problem, after expanding, we had terms like \(t^3\), \(t^2\), and constants (terms without any variable).

For the polynomial \((t^3 - 3t^2 + 3t - 1) + (at^3 + 3at^2 + 3at + a)\), we group terms as:- \((t^3 + at^3)\): both terms are cubes.- \((-3t^2 + 3at^2)\): squared terms.- \((3t + 3at)\): linear terms.- \((-1 + a)\): constant terms.Combining like terms not only simplifies your work but is also necessary for identifying coefficients and determining properties like the degree of a polynomial.

Solving Equations

The final step often involves solving an equation that arises from your simplified expression. In algebra, solving equations entails finding the values of variables that make an equation true. Once we have combined like terms in a polynomial, solving the equation becomes more straightforward:

• Identify the equation you need to solve. Here, the coefficient of the highest-degree term must reduce to zero to meet the problem's condition (degree less than 3).
• For the expression \((1+a)t^3 + \ldots\), the requirement was \(1+a=0\).

Solving \(1+a=0\) is straightforward:1. Subtract 1 from both sides: \(a = -1\).This equation-solving process allows us to find that \(a = -1\), ensuring the expanded polynomial's degree is less than 3, fulfilling the requirements of the original problem. The process of setting and solving an equation is fundamental in many algebraic contexts, offering the necessary solution to make the statement true.