Question

mr-ms+ns-nr factor this expression completely is what it says

Answer 1

\[\large mr-ms+ns-nr=m(r-s)-nr+ns\]
\[\large =m(r-s)-(nr-ns)\]
Factor nr-ns the same way I factored mr-ms.

Answer 2

After that, it's just a straight road to factoring this equation/expression.

Answer 3

@shivraj

Answer 4

thnks

Answer 5

No worries mate.

Answer 6

That question tried to trick you in all sorts. You just had to rearrange the last two terms.
Could you show me your final answer please?

Answer 7

@shivraj you still there?

Answer 8

hi

Answer 9

hi

Answer 10

mr-ms+ns-nr factor this expression completely is what it says

Answer 11

Do you remember how to use the distributive property of multiplication with respect to addition: $$ab+ac=a(b+c)$$

Answer 12

yes

Answer 13

Use it here:$$mr-ms$$ and here $$ns-nr $$

Answer 14

(r-s)(m-n)

Answer 15

Note: r-s is the opposite of s-r so $$(n-m)(s-r)$$ is also an acceptable answer

Answer 16

@shivraj any questions?

Answer 17

no sir

Answer 18

Kurt used the rule add 4,subtract 1 to generate a pattern. The first term in his pattern is 5. Which number could be in Kurt's pattern?