Chapter 8.7, Problem 45PPE

To determine

To find: A pair of factors for the given number 99 by using the difference of two squares.

Answer to Problem 45PPE

The factors of 99 are 9 and 11, so $99=9\cdot 11$ .

Explanation of Solution

Given information:

The given number is 99.

Formula used:

The factors of the difference of squares of two numbers a and b are: $a^2-b^2=\left(a-b\right)\left(a+b\right)$ .

Calculations:

The given number is 99 that may be written as

  $99=100-1$

  $99={\left(10\right)}^2-{\left(1\right)}^2$

Difference of two squares

  $99=\left(10-1\right)\left(12+1\right)$$\because a^2-b^2=\left(a-b\right)\left(a+b\right)$

  $99=\left(9\right)\left(11\right)$ .

Thus, the factors of 99are 9 and 11, so $99=9\cdot 11$ .

Conclusion:

The factors of 99 are 9 and 11, so $99=9\cdot 11$ .