不可分解多项式分析

Frage

因式分解：a^3b-ab^3+a^2+b^2+1

Gedanke

Thinking through possible factorization

I’m examining the expression $a^3 b - a b^3 + a^2 + b^2 + 1$ , grouping terms to factor by parts, and then considering using $(a + b + 1)$ to simplify or align terms.

Testing for symmetry

I’m working through the polynomial $a^3 b - a b^3 + a^2 + b^2 + 1$ , checking for factors like $(a + b + 1)$ and evaluating if $a = -b - 1$ is a root.

Piecing together polynomial terms

I substituted $a = -b - 1$ into $a^3 b - a b^3 + a^2 + b^2 + 1$ , combined polynomial terms, and simplified to $-2b^3 -2b^2 + b +1$ .

Final simplification

I’m noticing that combining terms from $a = -b - 1$ substitution resulted in $-2b^3 - b^2 + b + 2$ . It’s not trivially zero, suggesting further factorization might be complex.

Symmetrizing the expression

I’m exploring symmetrical factorization by reworking $a^2 + b^2 + 1$ into $(a + b)^2 + 1 - 2ab$ , and assessing the applicability of known polynomial identities.

Testing various methods

I’m piecing together symmetrical and partial groupings or known factorization patterns, like $a^3 + b^3 + c^3 - 3abc$ , to simplify the expression.

Examining integer roots

I'm contemplating a simpler method by parsing the polynomial as $b a^3 + a^2 - b^3 a + (b^2 + 1)$ to identify potential rational roots.

Testing rational roots

OK, let me see: I checked $a = b$ and $a = -b$ using the rational root theorem. Found some interesting zeroes but no general zero for all $b$ .

Piecing together

I tried $a = -1$ and found $b^3 + b^2 - b + 2$ . This suggests no polynomial factorization with rational coefficients. Exploring other factorization patterns is the next step.

Considering factorization

I’m exploring if $(a - b)$ or $(a + b)$ is a factor by substituting $a = b$ and $a = -b$ , but neither results in zero for all $b$ .

Trying another way

I'm thinking about adding and subtracting $2ab$ to $a^2 + b^2 + 1$ and then grouping terms to rewrite $ab(a - b)(a + b)$ as a sum of squares, but progress is tentative.

Reordering terms

I’m starting to see how to express $a^2 + b^2 + 1$ in a product form, possibly over complex numbers. Real polynomial factorization seems tricky, so complex numbers might be the key.

Pondering polynomial structure

I'm thinking about factoring the polynomial into a quadratic and a linear term in $a$ and $b$ . Polynomial long division will check if $\alpha a + \beta b + \gamma$ is a factor.

Adjusting my strategy

I’m thinking about integer values for $\alpha, \beta, \gamma$ such that $\alpha a + \beta b + \gamma$ makes the polynomial zero for all $b$ . This approach might simplify the polynomial.

Testing specific forms

I’m examining simpler expressions like $a + b - 1$ to spot factors that might simplify the polynomial. Consistent substitution and calculation of terms are progressing efficiently.

Revisiting roots

I’m mapping out simpler forms like $a - b + 1$ and substituting, but the polynomial still doesn't simplify to zero for all $b$ . No factor found here.

Testing simpler structures

I'm piecing together forms like $a - b + 1$ and $a - b$ , but they still don't make the polynomial zero for all $b$ . No factor found yet.

Testing potential simplifications

I tried $(a - b - 1)$ and $(a + 2b + 1)$ , but these forms don't simplify the polynomial to zero for all $b$ . It’s interesting to see that the polynomial remains complex.

Testing factorization methods

I'm exploring various partial factorizations and groupings like $(a^3 b + a^2) + (-a b^3 + b^2) + 1$ , but they don't simplify the polynomial to zero for all $b$ .

Evaluating irreducibility

I’m noting that the polynomial might be irreducible over real polynomials with rational coefficients. The expression seems complex, and complex domain factorization might be the answer.

Considering complex factorization

I’m mapping out complex factorization, assuming a linear factor in $a$ expressed through $b$ , and then solving for $a$ in terms of $b$ to identify any complex roots.

Including substitution

I’m substituting $a = x b$ into the cubic polynomial in $a$ to simplify and see if a linear factor emerges through this method.

Transforming variables

OK, let me see. I’m grouping terms in powers of $b$ , using $y = b^2$ to turn the expression into a quadratic in $y$ . Now, I'm seeking if this simplification aids in factorization.

Assessing degrees

I’m looking at the polynomial's total degree of 4 for each $a^3 b$ and $a b^3$ . This suggests the polynomial might be irreducible and not suitable for symmetrical factorization.

Thought about 因式分解，持续 1m 53s

Antwort

下面这道多项式

a^3 b \;-\; a\,b^3 \;+\; a^2 \;+\; b^2 \;+\; 1

很多同学在试图“拆”或“凑”时会怀疑它是不是能因式分解成较低次的因子。下面简单分析一下常见的分解思路，就会发现它其实在有理系数(甚至实系数)范围内是不可约的，换句话说，它无法再分解成次数更低的多项式相乘。

1. 尝试常见的提取公因式或分组

提取公因式：
前两项可以提取 $ab$ ：
$a^3 b - a b^3 \;=\; ab\,(a^2 - b^2) \;=\; ab\,(a-b)(a+b).$
于是原式可写成
$ab\,(a-b)(a+b)\;+\;a^2\;+\;b^2\;+\;1.$
但后面那团 $a^2 + b^2 + 1$ 并没有和前面的乘积形成明显的“配方”去进一步提取或合并，因而这种分组并没有带来可行的因式分解。
其它分组尝试：
- $\bigl(a^3 b + a^2\bigr) + \bigl(-a b^3 + b^2\bigr) + 1$ ,
- $\bigl(a^3 b + b^2\bigr) + \bigl(-a b^3 + a^2\bigr) + 1$ ,
  以及引入正负 $2ab$ 等“凑方”手段，都无法凑成常见的可分解结构(如 $(a+b)^2$ , $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ 等)，最终都卡住。

2. 检查简单的线性因子 $(\alpha a + \beta b + \gamma)$

若多项式能分解成两因子，其中一个是一次式 $\alpha a + \beta b + \gamma$ ，那么把它设为零后代入原式，就应对所有 $b$ 恒等为 0。这常常可以用来快速判断“有没有诸如 $(a \pm b \pm 1)$ 之类”的因子。

检验 $(a + b + 1)$ 为因子： 令 $a = -b - 1$ 代入，结果并不恒为零;
检验 $(a + b - 1)$ , $(a - b + 1)$ , $(a - b - 1)$ 等类似形式，代入后也都不恒为零。

更一般的整系数(或有理系数)一次式也都可以类似地代入排查，结论同样是否定。

3. 结论：在实(有理)系数下不可分

从上面的常规方法(分组、配方、尝试线性因子)都无果来看，这个四元混合(最高项是四次，且不齐次)的多项式并没有简单的分解形式。进一步若去做“将其视作关于 $a$ 的三次多项式，试图求根”的过程，也看不出会有一个 $\text{(线性因子)}\times \text{(二次多项式)}$ 的整(有理)系数分解。综合可判断：

在实(有理)系数范围内， $\;a^3 b - a b^3 + a^2 + b^2 + 1$ \ 是不可约的，
即无法分解成次数更低的实(有理)多项式相乘。

如果题目仅限于“实系数多项式因式分解”，那么它就已经是“分解结束”了(其实就是无法再分)。有时在更高等的课程里，允许用到复数根或更一般的代数扩张，那么它或许能拆成涉及复系数的因子，但那已经超出了通常“因式分解题目”的范畴。

因此，通常的结论就是：

\boxed{a^3 b - a b^3 + a^2 + b^2 + 1 \text{ 在实(有理)系数下不可再分解。}}