In a polynomial $$p\left( x\right) =x^{2}+x+41$$ put different values of x and find p(x). Can you conclude after putting different values of x that p(x) is prime for all. Is x an element of N? Put x=41 in p(x). Now what do you find?

Question

In a polynomial $$p\left( x\right) =x^{2}+x+41$$ put different values of x and find p(x). Can you conclude after putting different values of x that p(x) is prime for all. Is x an element of N? Put x=41 in p(x). Now what do you find?

Vedantu Content Team · Accepted Answer

- Hint: In this question it is given that $$p\left( x\right) =x^{2}+x+41$$ is a polynomial, we have to find the value of p(x) for different values of x and check whether the given polynomial is prime for all or not and also we have to find the value of p(x), when x=41. So to find the solution we need to know that a prime number is a natural number which only has two factors, one is 1 and another one is the number itself. So by using this information we find the obtained number is prime or not.

Complete step-by-step solution -
Given polynomial,
$$p\left( x\right) =x^{2}+x+41$$.........................(1)
Now we are going to put some of the values of x.
If x=0,
$$p\left( 0\right) =0^{2}+0+41=41$$
Which is a prime number.
When, x=1,
$$p\left( 1\right) =1^{2}+1+41$$
     $$=1+1+41$$
     $$=43$$
Which is a prime number.
When x=2,
$$p\left( 2\right) =2^{2}+2+41$$
     $$=4+2+41$$
     $$=47$$
47 is also a prime number.
When x=3,
$$p\left( 3\right) =3^{2}+3+41$$
     $$=9+3+41$$
     $$=53$$, is a prime number.
So from the above we have observed that for the above values of x, p(x) is a prime number.
Now let us check for x= 41.
$$\therefore p\left( 41\right) =(41)^{2}+41+41$$
      $$=41\times 41+41+41$$
      $$=1681+41+41$$
      $$=1763$$
Now if we divide 1763 by 41 then we get 43,
Therefore we can write,
$$p\left( 41\right) =1763=41\times 43$$
Which implies, apart from 1 and the number itself p(41) has two other factors, hence p(41) is not a prime number.
Also we can say that p(x) is not a prime number for all values of x.

Note: While solving this question you can also take the other values of x, but since the question is asking to find whether the polynomial p(x) is prime for every value of x. So that is why we put x=41 in order to show a contradictory statement, i.e, p(41) is not a prime number.