Question

Using the Horner’s method find the values of f(4) and f'(4) for the polynomial
f(x) = x^4 +2x^3 −x^2 +1.

Answer 1

ANSWER: $f(4)=369, f'(4)=344$

EXPLANATION:

$f(x)=a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=x^4+2x^3-x^2+1$

To calculate $f(4)$ , we represent the polynomial $f$ in the form $f(x)=(x-4)(b_3x^3+b_2x^2+b_1x+b_0)+ r$ , using the Horner's method.

${ b }_{ 3 }={ a }_{ 4 },\quad { b }_{ k-1 }=4{ b }_{ k }+a_{ k },\quad k=3,2,1,\quad r=4{ b }_{ 0 }+{ a }_{ 0 }$

Thus, we obtain ${ b }_{ 3 }=1\rightarrow { b }_{ 2 }=4\cdot 1+2=6\quad \rightarrow { b }_{ 1 }=4\cdot 6-1=23\rightarrow$

$\rightarrow { b }_{ 0 }=4\cdot 23+0=\rightarrow r=4\cdot 92+1=369, r=f(4)$

Similarly: $f'(x)={ 4x }^{ 3 }{ +6x }^{ 2 }-2{ x }^{ \ \ }=a_3x^3+a_2x^2+a_1x+a_0=$ $=(x-4)(b_2x^2+b_1x+b_0)+ r$

${ b }_{ 2 }=4\rightarrow { b }_{ 1}=4\cdot4+6=22\quad \rightarrow { b }_{ 0 }=4\cdot 22 -2=86\rightarrow$

$=\rightarrow r=4\cdot 86+0=344, r=f'(4)=344$