把开头乘上去的 x^2-1 也就是 (x-1)(x+1) 约去，就有. x^4+x^2+1=(x^2-x+1)(x^2+x+1)\\\\ 右端这两个二次式在实数范围已经无法分解了，因为它们 判别式 为负。这样就完成了工作。

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2008年9月7日 · x^4+x^2+1 =(x^4+2x^2+1)-x^2 =(x²+1)²-x² =[(x²+1)+x][(x²+1-x)] =(x²+x+1)(x²-x+1)

(x^2+x+1)^2 = x^4+x^2+1 by Freshmen's Dream. On the other hand, let f be a polynomial of degree 4 over \mathbb F_2, which has no roots. Then f is either irreducible or decomposes …

2015年5月10日 · How do you factor x4 + x2 + 1? x4 +x2 + 1 = (x2 + x + 1)(x2 −x + 1) To find this, first notice that x4 +x2 + 1> 0 for all (real) values of x. So there are no linear factors, only …

Identify like terms in the expression. Add or subtract the coefficients of like terms. Rewrite the expression with combined terms. Simplify 4x+5−2x+3. Combine the like terms (4x and −2x): …

2011年6月14日 · 用换元法将x^2看做一个整体t，变形为t^2-t+1形式，然后看判别式， b^2-4ac=（-1）^2-4*1*1=-3<0，所以说t^2-t+1=0这个方程是没有解的，所以也就不能进行分解因式！

Hence the factors of x 4 + x 2 + 1 are (x 2 + 1 + x) and (x 2 + 1 - x). Factorize : x4+x2+25. Factorize x^4 + x^2 + 1 Get the answer to this question and access a vast question bank that …

2011年11月30日 · 1/(x^4+x^2+1) = (Ax+B)/(x^2-x+1) + (Cx+D)/(x^2+x+1) A,B,C,D可以通过待定系数法确定 有理分式的积分方法相信书本上已经讲的很全面了，后面的你就自己算吧 仅供参考！

Your polynomial has roots $\frac{-1+\sqrt{3}}{2}$, $\frac{-1-\sqrt{3}}{2}$ as well as two complex cube roots of unity (since one of the factors of your polynomial is the 3rd cyclotomic polynomials).