Question

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

Answer 1

Hint: p(a) can be calculated by substituting x = a in the expression of p(x). Hence substitute x = 0, 1, 2 successively in the expression of p(x) 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 $p\left( x \right)={{x}^{3}}$
Substituting x= 0 in the expression of p(x), we get
$p\left( 0 \right)={{0}^{3}}=0$
Hence, we have p(0) = 0.
Calculating p(1):
We have $p\left( x \right)={{x}^{3}}$
Substituting x= 1 in the expression of p(x), we get
$p\left( 1 \right)={{\left( 1 \right)}^{3}}=1$
Hence, we have p(1) = 1.
Calculating p(2):
We have $p\left( x \right)={{x}^{3}}$
Substituting x=2 in the expression of p(x), we get
$p\left( 2 \right)={{\left( 2 \right)}^{3}}=8$
Hence, we have p(2) = 8.

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 $p\left( x \right)={{x}^{3}}={{x}^{3}}+0{{x}^{2}}+0x+0$ 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

Since the last carrydown is 0, we have $p\left( 0 \right)=0$.
For x =1:

Since the last carrydown is 1, we have $p\left( 1 \right)=1$
For x = 2:

Since the last carrydown is 8, we have $p\left( 2 \right)=8$