Waste less time on Facebook — follow Brilliant.
×

5th degree polynomial

Given the equation \(x^5+5(2x^3+x+2)=5x^2(x^2+2)\) has a real root in the form of \(a-\sqrt[5]{b}\) where \(a\) and \(b\) are real positive integers. Find \((a,b)\)

Note by William Isoroku
2 years, 9 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

\[x^5+5(2x^3+x+2)=5x^2(x^2+2)\\ x^5+10x^3+5x+10 = 5x^4+10x^2 \\ x^5 -5x^4+10x^3-10x^2+5x+10 = 0 \\ x^5 -5x^4+10x^3-10x^2+5x-1+1 +10 = 0 \\ (x-1)^5 = -11 \\ x-1 = - \sqrt [5] {11} \\ x = 1 - \sqrt [5] {11} \\ \Rightarrow \boxed{a = 1 \quad b = 11} \]

Chew-Seong Cheong - 2 years, 9 months ago

Log in to reply

Oh, so we have to know the \((x-1)^5\) factorization here right?

William Isoroku - 2 years, 9 months ago

Log in to reply

Yes, it is best that you remember first few rows of the Pascal triangle. They are actual the binomial expansion coefficients, which is given by \(\begin{pmatrix} n \\ r \end{pmatrix}\). For example, when \(n=5\), we have:

\[\begin{equation} \begin{split} (x-1)^5 & = \sum_{r=0}^5 {(-1)^r\begin{pmatrix} 5 \\ r \end{pmatrix}x^{5-r}} \\ & = \begin{pmatrix} 5 \\ 0 \end{pmatrix}x^5 - \begin{pmatrix} 5 \\ 1 \end{pmatrix}x^4 + \begin{pmatrix} 5 \\ 2 \end{pmatrix}x^3 - \begin{pmatrix} 5 \\ 3 \end{pmatrix}x^2 + \begin{pmatrix} 5 \\ 4 \end{pmatrix}x^1 - \begin{pmatrix} 5 \\ 5 \end{pmatrix}x^0 \\ & = x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1 \end{split} \end{equation} \]

Chew-Seong Cheong - 2 years, 9 months ago

Log in to reply

How you thought it will be (x-1)^5? Sir.

Ayush Verma - 2 years, 9 months ago

Log in to reply

Pascal triangle:

\[1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \\ 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1\\ 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1 \]

Chew-Seong Cheong - 2 years, 9 months ago

Log in to reply

@Chew-Seong Cheong Thanks

Ayush Verma - 2 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...