When we do the distributive, in the correct way, we must find 60x² on the first term. Therefore, let's multiply the terms on all options and see which one does not result in 60x².
Let's do the letter A first, we have
[tex](x\text{ })(60x\text{ })[/tex]When we do the distributive we would have
[tex]x\cdot60x=60x^2[/tex]As we wanted, then the letter A is appropriate. If the letter A is appropriate, let's test letter B
[tex]\begin{gathered} (2x\text{ })(30\text{ }) \\ \\ 2x\cdot30x=60x^2 \end{gathered}[/tex]As we expected.
For the letter C, we will have
[tex]\begin{gathered} (6x\text{ })(10x\text{ }) \\ \\ 6x\cdot10x=60x^2 \end{gathered}[/tex]Now the last one is the letter D.
[tex]\begin{gathered} (2x\text{ })(5x\text{ }) \\ \\ 2x\cdot5x=10x^2 \end{gathered}[/tex]That's not appropriate, the correct would be 60x², therefore, the inappropriate beginning is the letter D.
