Expand x^10+2x^9-3x^7-6x^6+3x^4+6x³-x-2 in powers
Expand $x^{10}+2x^9-3x^7-6x^6+3x^4+6x^3-x-2$ in powers of $(x-1)$ and show $x=1$ is a zero of order 3.
Explanation of Method
Rather than computing three derivatives by hand (error-prone for a degree-10 polynomial), grouping the eight terms into four pairs that each share a common factor $(x+2)$ reveals a clean, exact factorization. That factorization immediately identifies the exact multiplicity of the root at $x=1$ and gives the leading Taylor coefficient directly.
- Group terms in pairs to factor out $(x+2)$.
- Recognize the remaining factor as a perfect cube $(x^3-1)^3$.
- Read off the multiplicity of the root $x=1$ and the leading $(x-1)^3$ coefficient.
Step 1 · Group and factor
Let $f(x)=x^{10}+2x^9-3x^7-6x^6+3x^4+6x^3-x-2$. Group into four pairs: $$ f(x) = x^9(x+2) - 3x^6(x+2) + 3x^3(x+2) - 1\cdot(x+2) = (x+2)\left(x^9-3x^6+3x^3-1\right). $$
Step 2 · Recognize a perfect cube
With $a=x^3$: $a^3-3a^2+3a-1=(a-1)^3$ (binomial expansion of $(a-1)^3$). So $$ x^9-3x^6+3x^3-1 = (x^3-1)^3 = \bigl[(x-1)(x^2+x+1)\bigr]^3 = (x-1)^3(x^2+x+1)^3. $$ Therefore: $$ f(x) = (x+2)(x-1)^3(x^2+x+1)^3. $$
Step 3 · Multiplicity and leading coefficient
Since $x+2$ and $x^2+x+1$ do not vanish at $x=1$ (values $3$ and $3$, respectively), $(x-1)^3$ is the exact power of $(x-1)$ dividing $f(x)$ — so $x=1$ is a zero of order exactly $3$. Writing $t=x-1$ (so $x+2=t+3$ and $x^2+x+1=t^2+3t+3$): $$ f(x) = (t+3)\,t^3\,(t^2+3t+3)^3. $$ At $t=0$: the coefficient multiplying $t^3$ is $(0+3)(0^2+3\cdot0+3)^3=3\cdot27=81$. So the expansion begins $$ f(x) = 81(x-1)^3 + (\text{terms of degree} \ge 4 \text{ in } (x-1)), $$ confirming $f(1)=f'(1)=f''(1)=0$ and $f'''(1)=3!\cdot81=486\ne0$.
Summary Table
| Quantity | Value |
|---|---|
| Exact factorization | $(x+2)(x-1)^3(x^2+x+1)^3$ |
| Multiplicity at $x=1$ | $3$ |
| Leading coefficient of $(x-1)^3$ | $81$ |
| $f'''(1)$ | $486$ |
Additional Notes
- The grouping trick here — splitting eight terms into four pairs each sharing the factor $(x+2)$ — is far more reliable than computing three successive derivatives of a degree-10 polynomial by hand, where a single arithmetic slip can produce a spurious nonzero (or spurious zero) value.
- The remaining factor's recognition as $(x^3-1)^3$ is the key insight: the coefficients $1,-3,3,-1$ are exactly the binomial coefficients of $(a-1)^3$, immediately revealing the triple structure without any derivative computation at all.
- Full expansion in powers of $(x-1)$ to all ten terms would require expanding $(t+3)(t^2+3t+3)^3$ (a degree-7 polynomial in $t$, consistent with $3+7=10$ matching the original degree); the leading term $81(x-1)^3$ found here is the one relevant to establishing the order of the zero.