Chapter 8.7, Problem 57PPE

a.

To determine

To calculate: factorisethe equation by difference in square method.

Answer to Problem 57PPE

  $\left(4+9n^2\right)\left(2+3n\right)\left(2-3n\right)$

Explanation of Solution

Given:The expression is $16-81n^4$

Calculation:difference in square formula is givenby $a^2-b^2=\left(a+b\right)\left(a-b\right)$

  $\begin{array}{l}\Rightarrow 16-81n^4=\left(4^2-{\left(9n^2\right)}^2\right)=\left(4+9n^2\right)\left(4-9n^2\right)\\ =\left(4+9n^2\right)\left(2+3n\right)\left(2-3n\right)\end{array}$

b.

To determine

To show: The characteristics that make the above calculation possible.

Explanation of Solution

Given:The binomial given $16-81n^4$

Reason:According to the difference between square theorem the quadratic equation may be written as a difference between the two squares such that it is factorize into two binomials where one is the sum and other is the difference of the roots of the squares. This theorem can be used

until there is a difference between the squares.

Here in the above equation 16 is the square of 4 while 4 is the square of 2 and 81 is the square of 9 while 9 is the square of 3, same with the case of n.

Hence this characteristic makes the above solution possible.

c.

To determine

To calculate: Thebinomial that can be factoredthe equation by difference in square method.

Answer to Problem 57PPE

  $\left(16+25t^2\right)\left(4-5t\right)\left(4+5t\right)$

Explanation of Solution

Given:The expression is $216-625t^4$

Calculation:difference in square formula is givenby $a^2-b^2=\left(a+b\right)\left(a-b\right)$

  $\begin{array}{l}\Rightarrow 16-625t^4=\left({16}^2-{\left(25t^2\right)}^2\right)=\left(16+25t^2\right)\left(16-25t^2\right)\\ =\left(16+25t^2\right)\left(4-5t\right)\left(4+5t\right)\end{array}$