Question

If \[p(x) = {x^3} - {x^2} + 3x + 4\] then find \[p(1)\] and \[p( - 2)\]?

Answer 1

Hint: We have a polynomial of degree 3 and it is called a cubic polynomial. We can find \[p(1)\] and \[p( - 2)\] easily. To find \[p(1)\] all we need to put the value of ‘x’ equal to 1 in the given cubic equation. Similarly to find \[p( - 2)\] we need to put the value of ‘x’ equal to $-2$ in the given equation. After simplification we will have the required result.

Complete step by step answer:
Given, \[p(x) = {x^3} - {x^2} + 3x + 4\].
Now we need to find \[p(1)\] and \[p( - 2)\].
Now to find \[p(1)\], put \[x = 1\] in the given cubic polynomial that is \[p(x) = {x^3} - {x^2} + 3x + 4\].
Then we have
 \[ \Rightarrow p(1) = {\left( 1 \right)^3} - {\left( 1 \right)^2} + 3\left( 1 \right) + 4\]
\[ = 1 - 1 + 3 + 4\]
On simplifying we have
\[ = 8 - 1\]
\[ = 7\]
Thus we have \[ \Rightarrow p(1) = 7\].
Now to find \[p( - 2)\], put \[x = - 2\] in the given cubic polynomial that is \[p(x) = {x^3} - {x^2} + 3x + 4\]
Then we have
\[ \Rightarrow p( - 2) = {\left( { - 2} \right)^3} - {\left( { - 2} \right)^2} + 3\left( { - 2} \right) + 4\]
\[ = - 8 - 4 - 6 + 4\]
On simplifying we have
\[ = - 18 + 4\]
\[ = - 14\]
Thus we have \[ \Rightarrow p( - 2) = - 14\].
Hence \[ \Rightarrow p(1) = 7\] and \[ \Rightarrow p( - 2) = - 14\].

Note: We know that \[{( - a)^n} = {a^n}\] where n is an even number. We know that if we have a negative number which has a power which is an even number then we will have a positive number as a result. Similarly we know that \[{( - a)^m} = - {a^m}\] where m is an odd number. We know that if we have a negative number which has a power which is an odd number then we will have a negative number as a result.