Question

Determine the three degree polynomial for which it holds p(0) = 0 and P(x)-P(x - 1) =x2​

Answer 1

Given :  three degree polynomial for which it holds p(0) = 0 and P(x)-P(x - 1) =x²

To Find : Polynomial

Solution:

three degree polynomial

P(x)  = ax³  + bx²  + cx + d

P(0) = 0

=>  0 +0  + 0 + d  = 0

=>d = 0

P(x)  =  ax³  + bx²  + cx

P(x)-P(x - 1) =x²

=> P(x)   =   P(x - 1)  + x²

x = 1

=> P(1) = P(0) + x²

=> P (1) =  1²

P(1)  =  a   + b   + c  = 1

P(0)   =   P( - 1)  + 0²

=> P(-1) = 0

=>   -a + b - c  = 0

=> b = a + c

a   + b   + c  = 1

=> 2(a + c) = 1

=> a + c = 1/2

=>b = 1/2

P(2) = P(1)  + 2²

=> p(2) = 5

=>  8a + 4b  +2c  = 5

=> 8a + 4(a + c) + 2c = 5

=> 8a + 4(1/2) + 2c = 5

=> 8a + 2c = 3

8a + 2c = 3

 a + c = 1/2

=>2a +2c  = 1

=> 6a  = 2

=> a = 2/6 = 1/3

 a + c = 1/2

=> c = 1/2 - 1/3  = 1/6

a = 1/3  , b = 1/2  c = 1/6

P(X)  = x³/3  + x²/2  + x/6

Learn More:

on dividing x3-3x2+x+2 by a polynomial g(x),the quotient and ...

brainly.in/question/3135153

Divide 9x^3+3x^2-5x+7 by (3x-1) write the quotient and remainder ...

brainly.in/question/6190809

Divide the polynomial (6x³ + 11x² - 10x - 7) by the binomial (2x + 1 ...

brainly.in/question/4743553

Answer 2

$\textbf{Given:}$

$\mathsf{P(x)-P(x-1)=x^2\;and\;P(0)=0}$

$\textbf{To find:}$

$\textsf{The three degree polynomial for which the given condition holds}$

$\textbf{Solution:}$

$\textsf{Consider,}$

$\mathsf{P(x)-P(x-1)=x^2}$

$\mathsf{P(x-1)-P(x-2)=(x-1)^2}$

$\mathsf{P(x-2)-P(x-3)=(x-2)^2}$

$\;\;\mathsf{.}$

$\;\;\mathsf{.}$

$\;\;\mathsf{.}$

$\;\;\mathsf{.}$

$\;\;\mathsf{.}$

$\mathsf{P(3)-P(2)=3^2}$

$\mathsf{P(2)-P(1)=2^2}$

$\mathsf{P(1)-P(0)=1^2}$

$\textsf{Adding the above the equations, we get}$

$\mathsf{P(x)-P(x-1)+P(x-1)-P(x-2)+P(x-2)-P(x-3)}$$\mathsf{+\;.\;.\;.+P(3)-P(2)+P(2)-P(1)+P(1)-P(0)}$

$\mathsf{=x^2+(x-1)^2+(x-3)^2+\;.\;.\;.\;.+3^2+2^2+1^2}$

$\implies\mathsf{P(x)-P(0)=1^2+2^2+3^3\;.\;.\;.+(x-1)^2+x^2}$

$\implies\mathsf{P(x)-0=\dfrac{x(x+1)(2x+1)}{6}}$

$\implies\mathsf{P(x)=\dfrac{(x^2+x)(2x+1)}{6}}$

$\implies\mathsf{P(x)=\dfrac{2x^3+x^2+2x^2+x}{6}}$

$\implies\mathsf{P(x)=\dfrac{2x^3+3x^2+x}{6}}$

$\implies\mathsf{P(x)=\dfrac{2x^3}{6}+\dfrac{3x^2}{6}+\dfrac{x}{6}}$

$\implies\boxed{\mathsf{P(x)=\dfrac{x^3}{3}+\dfrac{x^2}{2}+\dfrac{x}{6}}}$