根据本人之前在知乎发表的题解中已经求得 x^ {8}-1 的因式分解的结果，所以我们已经知道了该方程的8个根分别如下：x12=±1=± cos…

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2015年12月20日 · #=(x-1)(x+1)(x^2+1)(x^2+sqrt(2)x+1)(x^2-sqrt(2)x+1)# This is as far as we can go with Real coefficients. The remaining quadratic factors all have Complex zeros.

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x^8 - 1) Natural Language; Math Input; Extended Keyboard Examples Upload Random. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music…

Solving algebra fractions with a quadratic - GCSE maths level 7+ - hard! How do you factor (x8)−16 ? x8-81 Final result : (x4 + 9) • (x2 + 3) • (x2 - 3) Step by step solution : Step 1 :Trying to factor as a Difference of Squares : 1.1 Factoring: x8-81 Theory : A difference of two perfect ...

2015年10月8日 · I take the #a# value, #x^4#, and take the #b# value, 1, and plug them into the equation. If we wanted to check this, we would need to expand the answer that we have gained using FOIL (First Outer Inner Last). I will demonstrate the steps below: #(x^4 - 1)(x^4 + 1)# First: #x^8 = (x^4)(x^4)# Outer: #x^4 = (x^4)(1)# Inner: #-x^4 = (x^4)(-1)# Last:

x^{2}-x-6=0 -x+3\gt 2x+1 ; line\:(1,\:2),\:(3,\:1) f(x)=x^3 ; prove\:\tan^2(x)-\sin^2(x)=\tan^2(x)\sin^2(x) \frac{d}{dx}(\frac{3x+9}{2-x}) (\sin^2(\theta))' \sin(120)

Using the formula we get $$ x^8 -1$$ $$ = (x^2 +1 + \sqrt {2}x) (x^2 +1- \sqrt {2}x)(x^2 +1)(x+1)(x-1) $$

Equations : Tiger Algebra gives you not only the answers, but also the complete step by step method for solving your equations x^8-1 so that you understand better

This problem is called a "Difference of two Squares" since x^8 is actually a perfect square x^4*x^4: x^8-1 (x^4-1)(x^4+1) The SUM of two Squares does NOT factor like the Difference of Squares factors, so leave it alone. However, notice that the first factor x^4-1 DOES factor again: (x^2-1)(x^2+1)(x^4+1)

NCERT Solutions for Class 10 Maths Chapter 14.

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Factor the expression x^8-1. We can factor the polynomial x^8-1 using the rational root theorem, which guarantees that for a polynomial of the form a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 there is a rational root of the form \pm\frac{p}{q}, where p belongs to the divisors of the constant term a_0, and q belongs to the divisors of the leading ...

The first step of factorising an expression is to take out any common factors which the terms have. So if we were asked to factor the expression $ {x^2} + x $ , since $ x $ goes into both terms, we would write $ x(x + 1) $ .

We have to break the term x 8 − 1 in the product of its factors such that when multiplied, they give us the value x 8 − 1. Let us write power 8 as product of 4 and 2. ∴ The factors of x 8 − 1 are (x − 1); (x + 1); (x 2 + 1); (x 4 + 1).

2013年4月3日 · The question was about the number of roots of \(x^8-1\) in \(\mathbb{R}\) and in \(\mathbb{C}\). In the real numbers, we identified two roots: \(1\) and \(-1\). These are obviously also complex numbers, as their imaginary part (the \(b\) in the \(a+bi\) form) is just \(0\).

2023年4月7日 · The given algebraic expression is $x^8-1$. To do: We have to factorize the expression $x^8-1$. Solution: Factorizing algebraic expressions: Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution.

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