Chapter 8.7, Problem 6CYU

To determine

To find: That the given polynomial is a difference of squares, write yes or no. If yes then factor it.

Answer to Problem 6CYU

The polynomial $r^2-9t^2$ is a difference of two squares, so it is factorable. The factored polynomial is $r^2-9t^2=\left(r-3t\right)\left(r+3t\right)$ .

Explanation of Solution

Given information:

The given polynomial is $r^2-9t^2$ .

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 $P=r^2-9t^2$ .

The polynomial can be written as $P=r^2-{\left(3t\right)}^2$ , i.e., difference of two squares.

Since, the given polynomial $r^2-9t^2$ is a difference of two squares, so it can be factorized.

The factors of polynomial P are

  $P={\left(r\right)}^2-{\left(3t\right)}^2$

  $\Rightarrow P=\left(r-3t\right)\left(r+3t\right)$

Thus, the given polynomial $r^2-9t^2$ is a difference of two squares, so it can be factorized.

The factored polynomial is $P=r^2-9t^2=\left(r-3t\right)\left(r+3t\right)$ .

Conclusion:

The polynomial $r^2-9t^2$ is a difference of two squares, so it is factorable. The factored polynomial is $r^2-9t^2=\left(r-3t\right)\left(r+3t\right)$ .