Chapter 8.7, Problem 3CYU

To determine

To find: That the given polynomial is a difference of squares, if yes then to factor it.

Answer to Problem 3CYU

No, the polynomial $9a^2+1$ is a prime polynomial.

Explanation of Solution

Given information:

The given polynomial is $9a^2+1$ .

Concept used:

The difference of two squares can be factored as $a^2-b^2=\left(a-b\right)\left(a+b\right)$ .

Calculations:

The given polynomial P (say) is $P=9a^2+1$ .

The polynomial can be written as $P={\left(3a\right)}^2+{\left(1\right)}^2$ , i.e., not the difference of two squares.

Therefore, the polynomial $P=9a^2+1$ cannot be factorized.

Hence, the polynomial $9a^2+1$ is a prime polynomial.

Conclusion:

The polynomial $9a^2+1$ cannot be written as the difference of two squares, it is a prime polynomial.