Question

Find the value of p(0), p(1) and p(2) where p(t) is given by
$p\left( t \right)=2+t+2{{t}^{2}}-{{t}^{3}}$

Answer 1

Hint: p(a) can be calculated by substituting x = a in the expression of p(x). Hence substitute t = 0, 1, 2 successively in the expression of p(t) to get the value of p(0), p(1) and p(2) respectively.

Complete step-by-step answer:
Alternatively, use synthetic division to get the value of p(0), p(1) and p(2).

Calculating p(0):
We have $p\left( t \right)=2+t+2{{t}^{2}}-{{t}^{3}}$
Substituting t= 0 in the expression of p(t), we get
$p\left( 0 \right)=2+0+2{{\left( 0 \right)}^{2}}-{{0}^{3}}=2+0+0-0=2$
Hence, we have p(0) = 2.
Calculating p(1):
We have $p\left( t \right)=2+t+2{{t}^{2}}-{{t}^{3}}$
Substituting t= 1 in the expression of p(t), we get
$p\left( 1 \right)=2+1+2{{\left( 1 \right)}^{2}}-{{1}^{3}}=2+1+2-1=4$
Hence, we have p(1) = 4.
Calculating p(2):
We have $p\left( t \right)=2+t+2{{t}^{2}}-{{t}^{3}}$
Substituting t= 2 in the expression of p(t), we get
$p\left( 2 \right)=2+2+2{{\left( 2 \right)}^{2}}-{{2}^{3}}=2+2+8-8=4$
Hence, we have p(2) = 4.

Note: Alternative method: Synthetic division: Best method.
In this method, we start by writing coefficients of the polynomial in order from the highest degree to the constant term. If in between some degree terms are missing we set their coefficient as 0.
Hence $p\left( t \right)=2+t+2{{t}^{2}}-{{t}^{3}}=-{{t}^{3}}+2{{t}^{2}}+t+2$ will be written as

 Now each the point which has to be substituted(say x= 0) is written as follows

0 is placed below the first term

Now the terms under the same column are added. The sum is then multiplied with the root, and the product is written under the coefficient of the next term.
Hence, we have

Continuing in this way we have the following

Since the last sum is 2, we have $p\left( 0 \right)=2$. This method also tells you what the quotient will be when p(t) is divided by t-0. Here it will be $-1\left( {{t}^{2}} \right)+2\left( t \right)+1=-{{t}^{2}}+2t+1$
Similarly creating tables for t = 1 and t = 2, we get

Since the last sum is 4, we have $p\left( 1 \right)=4$

Since the last sum is 4, we have $p\left( 2 \right)=4$