Question

What is the product of the quantity m squared minus m minus 6 all over 8 times p • 2 times p all over 2 times m plus m squared?

Answer 1

m squared : \(m^2\)

minus m : \(-m\)

minus 6 : \(-6\)

all over 8 times p: \(\large \frac{m^2 - m - 6}{8 \times p}\)

2 times p : \(2p\)

all over 2 times m plus m squared means : \(\large \frac{2p}{2 m + m^2}\)

Answer 2

So this becomes:

\[\large \frac{m^2 - m - 6}{8 \times p} \times \frac{2p}{2 m + m^2} = ??\]

Answer 3

Mhmm

Answer 4

Can you factorize it :

\[m^2 - m - 6\]

Answer 5

(m - 3)(m + 2)

Answer 6

Yes.. great..

Answer 7

Now what? :3

Answer 8

Can you factor out m from here:

\[2m + m^2\]

Answer 9

m(2 + m)

Answer 10

Oh I should cancel out the 2 + m

Answer 11

Now your work is to cancel things in numerator and denominator..

Answer 12

Yes cancel the things you can cancel..
And tell me what are you left with ??

Answer 13

(m - 3) 2p
_____________ * _____________
8p m

Answer 14

Yes..

You cancel p and 2 by 8 also..

Answer 15

m - 3
_____________
4m

Answer 16

Good Job...

Answer 17

Would it be, m ≠ −2, 0 and p ≠ 0 or m ≠ 0, 2 and p ≠ 0

Answer 18

It will be simply m not equal to 0..

Answer 19

I mean m ≠ −2, 0 or m ≠ 0, 2

Answer 20

If you are looking at the original equation then:

\[p \ne 0, m \ne 0, m \ne -2\]