Question

Decompose \(1 /\left\\{(\mathrm{x}+1)\left(\mathrm{x}^{4}+2 \mathrm{x}^{3}+3 \mathrm{x}^{2}+1\right)\right\\}\) into partial functions .

Short answer

The original function \(\frac{1}{(x+1)(x^4+2x^3+3x^2+1)}\) is already an irreducible fraction, and its decomposition into partial fractions is simply \(\frac{1}{x^4+2x^3+3x^2+1}\), with the coefficients A, B, C, and D all equalling 0.

Step-by-step solution

Factor the denominator

First, we need to factor the denominator. In this case, the denominator is already factored as \((x+1)(x^4+2x^3+3x^2+1)\).

Setup partial fractions

Next, we will set up the partial fractions. Since one of the factors in the denominator is a linear term \((x+1)\) and the other factor is a quartic polynomial \((x^4+2x^3+3x^2+1)\), we will have the following form: \[\frac{1}{(x+1)(x^4+2x^3+3x^2+1)} = \frac{A}{x+1} + \frac{Bx^3+Cx^2+Dx+E}{x^4+2x^3+3x^2+1}\]

Clear the denominator

Now, we need to clear the denominator by multiplying both sides of the equation by the original denominator: \((x+1)(x^4+2x^3+3x^2+1)\): \[1 =A(x^4+2x^3+3x^2+1) + (Bx^3+Cx^2+Dx+E)(x+1)\]

Expand and collect terms

Expand the equation and collect like terms: \[1 = A x^4 + (2A+B) x^3 + (3A+C) x^2 + (A+D) x + (A+E)\]

Equating coefficients

Now, we will equate the coefficients of the corresponding terms on both sides of the equation. This gives us the following system of linear equations: \[ \begin{cases} A = 0 \\ 2A + B = 0 \\ 3A + C = 0 \\ A + D = 0 \\ A + E = 1 \end{cases} \]

Solve the system of equations

Solving this system of linear equations, we get: \[ \begin{cases} A = 0 \\ B = 0 \\ C = 0 \\ D = 0 \\ E = 1 \end{cases} \]

Rewrite the function

Now, we can rewrite the original function using the values of A, B, C, D, and E in the partial fractions: \[\frac{1}{(x+1)(x^4+2x^3+3x^2+1)} = \frac{0}{x+1} + \frac{0x^3+0x^2+0x+1}{x^4+2x^3+3x^2+1}\] \[\frac{1}{(x+1)(x^4+2x^3+3x^2+1)} = \frac{1}{x^4+2x^3+3x^2+1}\] So, the original function has been decomposed into partial fractions. Note that since A, B, C, and D all equal 0, the original function is an irreducible fraction. Therefore, the original function was already in the simplest form.

Key concepts

Factoring Polynomials

Understanding how to factor polynomials is crucial when tackling exercises involving partial fraction decomposition. Factoring polynomials involves breaking down a polynomial into simpler 'factors' that, when multiplied together, give back the original polynomial.

For instance, in the exercise provided, the denominator of the function \( (x+1)(x^4+2x^3+3x^2+1) \) is already factored. However, it's essential to recognize when a polynomial can be further factored because it affects how we set up the partial fractions. If the quartic term could be factored into linear or quadratic factors, we would have additional terms in our partial fraction setup. As it stands, because it's already simplified and can't be factored further, we proceed with it as a whole.

System of Linear Equations

With partial fraction decomposition, we often encounter a system of linear equations. This occurs when we equate coefficients of corresponding terms to solve for unknowns in our fraction setups.

In this exercise, we ended up with a system where \( A, B, C, \) and \( D \) were all equal to zero after equating the coefficients, simplifying the function significantly. Systems of linear equations are pivotal in algebra and pre-calculus and getting comfortable with them will help solve various mathematical problems including those beyond partial fractions.

Irreducible Fraction

An irreducible fraction is a fraction that cannot be simplified further; the numerator and the denominator have no common factors. Throughout the decomposition of partial fractions, we aim to break down complex fractions into simpler components. However, as the solution reveals, sometimes our initial fraction is already irreducible, as was the case with the original function in the exercise.

This highlights the importance of understanding when a fraction is at its simplest form. Recognizing irreducible fractions saves time and effort by indicating that no further decomposition is necessary, as we found with the coefficients \(A = 0, B = 0, C = 0, D = 0,\) and \(E = 1\).

Pre-calculus

Mastering partial fraction decomposition is often part of a pre-calculus course where students get familiar with a wide range of topics preparing them for calculus. This subject includes polynomials, systems of equations, and irreducible fractions, all of which are foundational for understanding calculus concepts like integration and limits.

Pre-calculus is not just a set of random concepts; it's a carefully structured preparation where each topic, like the ones discussed through this exercise, builds on the last to ensure a deeper understanding of mathematics. It enables students to handle more advanced functions and their manipulations, essential for the study of continuous change in calculus.