Problem 64 - Precalculus

Question

Solve the system of linear equations. Find the values of \(A, B, C, D,\) and \(E\) such that the following equation is true: $$ \begin{array}{l} A x^{3}(x+1)+B x^{2}(x+1)+C x(x+1)+D(x+1)+E x^{4} \\\ \quad=4 x^{4}+x+1 \end{array} $$

Step-by-Step Solution

Answer

\(A = 0, B = 0, C = 0, D = 1, E = 4\)

Step 1: Expand both sides of the equation

First, expand the given expression on the left-hand side. We need to distribute each term individually. The expression becomes: \[ A x^4 + A x^3 + B x^3 + B x^2 + C x^2 + C x + D x + D + E x^4 \]

Step 2: Combine like terms

Group the terms with the same power of \(x\) together on the left-hand side:\[ (A + E) x^4 + (A + B) x^3 + (B + C)x^2 + (C + D) x + D \]

Step 3: Compare coefficients on both sides

On the right-hand side, the polynomial is given by:\[ 4x^4 + 0x^3 + 0x^2 + 1x + 1 \]By comparing the coefficients of like terms from both sides, we get:- Coefficient of \(x^4\): \( A + E = 4 \)- Coefficient of \(x^3\): \( A + B = 0 \)- Coefficient of \(x^2\): \( B + C = 0 \)- Coefficient of \(x\): \( C + D = 1 \)- Constant term: \( D = 1 \)

Step 4: Solve for D

From the constant term equation, we directly get:\[ D = 1 \]

Step 5: Solve for C

Using the equation \( C + D = 1 \):\[ C + 1 = 1 \]Solve for \( C \):\[ C = 0 \]

Step 6: Solve for B

Using the equation \( B + C = 0 \):\[ B + 0 = 0 \]Solve for \( B \):\[ B = 0 \]

Step 7: Solve for A

Using the equation \( A + B = 0 \):\[ A + 0 = 0 \]Solve for \( A \):\[ A = 0 \]

Step 8: Solve for E

Using the equation \( A + E = 4 \):\[ 0 + E = 4 \]Solve for \( E \):\[ E = 4 \]

Key Concepts

Polynomial Expansion

Polynomial expansion is the process of multiplying out expressions, ensuring each term in the first polynomial is distributed across all terms in the second polynomial. In our exercise, we are given the polynomial \[ A x^3(x+1) + B x^2(x+1) + C x(x+1) + D(x+1) + E x^4 \]and we must expand it. This process involves:

• Taking each term inside the parentheses and multiplying it by each term outside.
• Combining results into a new polynomial that shows each possible product of terms.

For example, for the term \( A x^3(x+1) \), we distribute to get: \( A x^4 + A x^3 \). This same principle applies to all terms, leading to a fully expanded expression where similarities or identical terms can be identified and grouped.

Comparing Coefficients

Comparing coefficients is a critical technique for solving polynomial equations where expressions on both sides of the equation must be identical. After expanding both sides, we look at the polynomial
\[(A+E)x^4 + (A+B)x^3 + (B+C)x^2 + (C+D)x + D \]
which should equal
\[4x^4 + 0x^3 + 0x^2 + 1x + 1\].
Here, the rule is straightforward:

• Match each power of \(x\) from both sides: the coefficients must equal each other.
• This provides a series of equations to solve for unknowns.

Take, for example, \((A+E)x^4\): This must match the coefficient of \(x^4\) on the right side, which is 4, creating the equation \(A + E = 4\). By systematically aligning each term this way, we derive specific equations that can be solved for the unknown values of \(A, B, C, D,\) and \(E\).

Combining Like Terms

Combining like terms is an essential step to simplify polynomial expressions, particularly after expansion. Each step in the original problem stressed merging terms with the same degree to simplify equations. In this instance:

• This involves taking each set of terms that have the same power of \(x\) and adding their coefficients together.
• For example, combining all terms in the expanded expression with \(x^4\) terms gives \((A+E)x^4\).

The process clarifies the polynomial’s structure by reducing it to a more straightforward form, enabling us to more easily compare each component to its respective parts on the other side of the equation. This capability is crucial as it establishes the foundation for correctly applying the method of comparing coefficients, bridging our expansion work into practical, solvable equations.