Chapter 8.3, Problem 45PPE

a.

To determine

To simplify: The given pair of products

Answer to Problem 45PPE

For $i)$

  $\begin{array}{l}\left(x+1\right)\left(x+1\right)=x^2+2x+1\\ 11.11=121\end{array}$

For $ii)$

  $\begin{array}{l}\left(x+1\right)\left(x+2\right)=x^2+3x+2\\ 11.12=132\end{array}$

For $iii)$

  $\begin{array}{l}\left(x+1\right)\left(x+3\right)=x^2+4x+3\\ 11.13=143\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}i)\\ \left(x+1\right)\left(x+1\right)\\ 11.11\\ \\ ii)\\ \left(x+1\right)\left(x+2\right)\\ 11.12\\ \\ iii)\\ \left(x+1\right)\left(x+3\right)\\ 11.13\end{array}$

Calculation:

For $i)$

  $\begin{array}{l}\left(x+1\right)\left(x+1\right)=x\left(x+1\right)+1\left(x+1\right)=x^2+x+x+1=x^2+2x+1\\ \\ 11.11=\left(10+1\right)\left(10+1\right)=10\left(10+1\right)+1\left(10+1\right)=100+10+10+1=121\end{array}$

For $ii)$

  $\begin{array}{l}\left(x+1\right)\left(x+2\right)=x\left(x+2\right)+1\left(x+2\right)=x^2+2x+x+2=x^2+3x+2\\ \\ 11.12=\left(10+1\right)\left(10+2\right)=10\left(10+2\right)+1\left(10+2\right)=100+20+10+2=132\end{array}$

For $iii)$

  $\begin{array}{l}\left(x+1\right)\left(x+3\right)=x\left(x+3\right)+1\left(x+3\right)=x^2+3x+x+3=x^2+4x+3\\ \\ 11.13=\left(10+1\right)\left(10+3\right)=10\left(10+3\right)+1\left(10+3\right)=100+30+10+3=143\end{array}$

b.

To determine

To find: Similarities between two answers in each pair.

Answer to Problem 45PPE

You get the same value if you replace the x by 10 in variable answer

Explanation of Solution

Given:

  $\begin{array}{l}i)\\ \left(x+1\right)\left(x+1\right)\\ 11.11\\ \\ ii)\\ \left(x+1\right)\left(x+2\right)\\ 11.12\\ \\ iii)\\ \left(x+1\right)\left(x+3\right)\\ 11.13\end{array}$

You get the same value if you replace the x by 10 in variable answer