Question

Consider the expression below. Assume the variable m represents an integer. 6m(3m + 21) Enter an expression in the box that uses the variable m and makes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.) 6m(3m + 21) = 9 Given that m represents an integer, is 6m(3m + 21) divisible by 9?

Answer 1

Answer:

$(a)\ 6m(3m + 21) = 9(2m^2 + 14m)$

(b) Yes, it is divisible by 9

Step-by-step explanation:

Given

$6m(3m + 21)$

Solving (a): Complete the blanks

$6m(3m + 21) = 9 [\ ]$

Expand the bracket

$6m(3[m + 7]) = 9 [\ ]$

$6m*3(m + 7) = 9 [\ ]$

Express 6m as 2m * 3

$2m*3*3(m + 7) = 9 [\ ]$

$2m*9(m + 7) = 9 [\ ]$

Rewrite as:

$9 * 2m(m + 7) = 9 [\ ]$

Multiply the bracket by 2m

$9 * (2m^2 + 14m) = 9 [\ ]$

Divide both sides by 9

$2m^2 + 14m = [\ ]$

Hence, the bracket will be filled with: $2m^2 + 14m$

So:

$6m(3m + 21) = 9(2m^2 + 14m)$

Solving (b): Is $6m(3m + 21)$ divisible by 9?

In (a), we have:

$6m(3m + 21) = 9(2m^2 + 14m)$

The leading factor "9" implies that the expression is divisible by 9