Answer :
Firstly look for the highest number which goes into both of them:
1 goes into both of them
3 goes into both of them and is the highest so we use this one.
Next write a 3 outside a set of brackets:
[tex]3(\ )[/tex]
Then, the point of these brackets is that you times everything inside them by 3 so in order to get the numbers inside, you divide both by 3.
[tex]27 \div 3 = 9[/tex]
[tex]12 \div 3 = 4[/tex]
So now put these back into the brackets and you're done:
[tex]3(9+4)[/tex]
Try to factorise [tex]18+12[/tex] now to make sure you understand.
1 goes into both of them
3 goes into both of them and is the highest so we use this one.
Next write a 3 outside a set of brackets:
[tex]3(\ )[/tex]
Then, the point of these brackets is that you times everything inside them by 3 so in order to get the numbers inside, you divide both by 3.
[tex]27 \div 3 = 9[/tex]
[tex]12 \div 3 = 4[/tex]
So now put these back into the brackets and you're done:
[tex]3(9+4)[/tex]
Try to factorise [tex]18+12[/tex] now to make sure you understand.
= 27 + 12
3 is common in both of them so
= 3 (9 + 4)
The answer is 3(9+4).
3 is common in both of them so
= 3 (9 + 4)
The answer is 3(9+4).
