Expand the polynomial x³-2x²+5x-7 in powers of (x-1)
Expand the polynomial $x^3-2x^2+5x-7$ in powers of $(x-1)$.
Explanation of Method
For a polynomial, the Taylor expansion about any point terminates exactly at the polynomial's degree, so it can be found either by evaluating derivatives at $x=1$, or equivalently by direct algebraic substitution $x=(x-1)+1$ and expanding. Both routes give the same finite expansion; substitution is a useful cross-check.
- Compute $f(1),f'(1),f''(1),f'''(1)$.
- Assemble the Taylor expansion $\sum f^{(n)}(1)(x-1)^n/n!$.
- Verify via direct substitution $x=(x-1)+1$.
Step 1 · Derivatives at $x=1$
With $f(x)=x^3-2x^2+5x-7$: $$ f(1)=1-2+5-7=-3, \quad f'(x)=3x^2-4x+5,\ f'(1)=4, $$ $$ f''(x)=6x-4,\ f''(1)=2, \qquad f'''(x)=6,\ f'''(1)=6. $$
Step 2 · Assemble the expansion
$$ f(x) = f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3 = -3+4(x-1)+(x-1)^2+(x-1)^3. $$
Step 3 · Verify by substitution
Let $y=x-1$, $x=y+1$: $$ (y+1)^3-2(y+1)^2+5(y+1)-7 = (y^3+3y^2+3y+1)-2(y^2+2y+1)+5y+5-7 $$ $$ = y^3+y^2+4y-3. $$ This matches Step 2's result exactly (with $4y$ vs. $4(x-1)$, etc.), confirming the expansion.
Summary Table
| Coefficient | Value |
|---|---|
| $(x-1)^0$ | $-3$ |
| $(x-1)^1$ | $4$ |
| $(x-1)^2$ | $1$ |
| $(x-1)^3$ | $1$ |
Additional Notes
- For any polynomial of degree $d$, the Taylor expansion about any center is exact and terminates after $d+1$ terms — there is no remainder or convergence question to worry about, unlike the transcendental-function expansions in Problems 249–254.
- The two methods (derivatives vs. direct substitution) are algebraically equivalent ways of computing the same binomial-theorem expansion; substitution is often faster for low-degree polynomials, while the derivative method generalizes immediately to non-polynomial functions.
- This is the natural warm-up for Problem 245, which asks the same kind of question for a much higher-degree polynomial, but adds the twist of identifying a repeated root via the vanishing of several leading coefficients.