Exercise 539Algebra › PolynomialsHigh schoolAdded 3 Nov 2026

Horner's scheme

#Horner scheme#evaluation#algorithm

Statement

Use Horner's scheme to evaluate $P(X)=2X^4-3X^3+X^2-5X+4$ at $X=2$ .

Hint

Repeatedly factor: $P(X)=(((2X-3)X+1)X-5)X+4$ .

Show solutionHide solution

Horner steps.
- Start: $b_4=2$ .
- $b_3=2\cdot2-3=1$ .
- $b_2=1\cdot2+1=3$ .
- $b_1=3\cdot2-5=1$ .
- $b_0=1\cdot2+4=6$ .

Conclusion. $P(2)=6$ .

Check. Direct: $2\cdot16-3\cdot8+4-10+4=32-24+4-10+4=6$ . Correct.

Remark. Horner's scheme requires $n$ multiplications and $n$ additions for degree $n$ , vs. $O(n^2)$ for naive evaluation.