Answer to Question #189632 in Algebra for Bansari

Question #189632

Evaluate the following polynomial expression using Horner’s rule. Assume the value of x

is 3.

𝑃(π‘₯) = 10x5 + 6x4+ 3x3 βˆ’ 6x2 + 8π‘₯ + 15

Show all the intermediate steps.


1
Expert's answer
2021-05-16T21:56:01-0400

Given, the polynomial p(x)=10x5+6x4+3x3βˆ’6x2+8x+15andx0=3Since, it is 5 degree polynomial.n=5Hornerβ€²srule,1.Setk=n2.Letbk=ak3.Letbkβˆ’1=akβˆ’1+bkx04.Letk=kβˆ’15.Ifkβ‰₯0then go to step 3Else EndThen,b5=a5=10b4=a4+x0b5=6+3Γ—10=36b3=a3+x0b4=3+3Γ—36=111b2=a2+x0b3=βˆ’6+3Γ—111=327b1=a1+x0b2=8+3Γ—327=989b0=a0+x0b1=15+3Γ—989=2982Therefore,P(3)=2982.\text{Given, the polynomial }p(x)=10x^5+6x^4+3x^3-6x^2+8x+15 and x_0=3\newline \text{Since, it is 5 degree polynomial.} n=5\newline Horner's rule,\newline 1. Set k = n\newline 2. Let b_k = a_k\newline 3. Let b_{k - 1} = a_{k - 1 }+ b_kx_0\newline 4. Let k = k - 1\newline 5. If k β‰₯ 0 \text{then go to step 3}\newline \text{Else End}\newline Then,\newline b_5=a_5=10\newline b_4 = a_4 + x_0b_5=6+3Γ—10=36 \newline b_3 = a_3 + x_0b_4=3+3Γ—36=111 \newline b_2 = a_2 + x_0b_3=-6+3Γ—111=327 \newline b_1 = a_1 + x_0b_2=8+3Γ—327=989 \newline b_0 = a_0 + x_0b_1=15+3Γ—989=2982 \newline Therefore,P(3)=2982.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment