---
title: "Express x³−5x²+6x+1 in Legendre polynomials"
subject: "Mathematics"
chapter: 4
key: "semester_2/mathematics:express-x3-5x2-6x-1"
source: https://www.bctengineer.com/q/semester_2/mathematics/express-x3-5x2-6x-1
site: BCT Engineering
---

# Express x³−5x²+6x+1 in Legendre polynomials

Mathematics · Chapter 4: Solution of Differential Equations in Series and Special Functions

- Asked in: 2073 Magh, 5 marks; 2068 Bhadra, 5 marks (asked 2 times)
- Video: [Express f(x) interms of Legendre's polynomials - Example Problem 3 | Engineering Mathematics](https://www.youtube.com/watch?v=ocqWh8n98v4) — Mathematics Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/express-x3-5x2-6x-1
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-4

## Question

Express in terms of Legendre's polynomials f(x) = x³ − 5x² + 6x + 1.

## Short answer

Express x³, x², x, 1 using P_0=1, P_1=x, P_2=(3x²-1)/2, P_3=(5x³-3x)/2. Substituting into x³-5x²+6x+1 gives (2/5)P_3(x) - (10/3)P_2(x) + (33/5)P_1(x) - (2/3)P_0(x).

## Answer

Legendre polynomials: P₀ = 1, P₁ = x, P₂ = (3x² − 1)/2, P₃ = (5x³ − 3x)/2.

#### Powers of x in terms of Pₙ
```text
x³ = (2P₃ + 3P₁)/5        (from P₃ = (5x³ − 3x)/2)
x² = (2P₂ + P₀)/3         (from P₂ = (3x² − 1)/2)
x  = P₁,   1 = P₀
```

#### Substitute in f(x) = x³ − 5x² + 6x + 1
```text
f = (2/5)P₃ + (3/5)P₁ − (5/3)(2P₂ + P₀) + 6P₁ + P₀
  = (2/5)P₃ − (10/3)P₂ + (3/5 + 6)P₁ + (1 − 5/3)P₀
  = (2/5)P₃ − (10/3)P₂ + (33/5)P₁ − (2/3)P₀
```

**Answer: x³ − 5x² + 6x + 1 = (2/5)P₃(x) − (10/3)P₂(x) + (33/5)P₁(x) − (2/3)P₀(x).**

Check at x = 1: 2/5 − 10/3 + 33/5 − 2/3 = 7 − 4 = 3, and f(1) = 1 − 5 + 6 + 1 = 3 ✓.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which polynomials are needed for a cubic?
   A. P1, P2 only
   B. P0, P1, P2, P3
   C. P0, P3 only
   D. P2, P3 only
   Answer: B. A degree 3 polynomial needs degrees 0 to 3. (Key point: Basis)

2. The coefficient of P1 comes from 3/5 (from x^3) plus:
   A. 1
   B. 5
   C. 6
   D. -6
   Answer: C. 6x = 6P1 so 3/5 + 6 = 33/5. (Key point: P1 coefficient)

3. The coefficient of P0 is:
   A. -2/3
   B. 2/3
   C. 1
   D. -5/3
   Answer: A. 1 - 5/3. (Key point: P0 coefficient)

4. At x = 1 all P_n = 1, so the coefficients sum to:
   A. 0
   B. f(1) = 3
   C. 1
   D. 13
   Answer: B. 1 - 5 + 6 + 1 = 3. (Key point: Check)

## Close variants set in other exams

- [Express 5x³+x in Legendre polynomials](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aexpress-5x3-x-legendre-polynomials) — Mathematics, chapter 4, asked 3×

## Related in Mathematics

- [Express x³−5x²+x+2 in Legendre polynomials](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aexpress-x3-5x2-x-2) — Mathematics, chapter 4, asked 3×