Today we skipped homework checking and went straight to the lesson, so it's probably best to use Mr. Cope's moodle key to check your homework before the homework quiz, whenever it may be.
We then began to simplify radicals to tie together past lessons with yesterday's lesson, all to today's lesson. Mr. Copes goal: for us to be able to simplify ⁵√64x¹¹y²⁷z⁵³ to 2x²y⁵z¹⁰ ⁵√2xy²z³ .
Mr. Cope skipped the formal definitions, for a more practical approach, by using one of his colleagues explanations of factoring, The Jailbreak Method. To use the Jailbreak method one needs a pair of the same number to "break out" of the radical, while the unpaired numbers remain under the radical.
He then linked this to todays lesson by explaining that when you factor out various types of radicals like ∛ ∜, instead of just looking for a pair, you look for the number thats written. In these cases, groups of 3's or 4's.
To simplify radicals, you must factor out the radicals, then look for the groups, indicated by the number before the radical. The number(s) that can complete the number needed "break out" of the radical, while the rest must remain under.
While the purple 2's form a group of three, the red 2 & 5 don't, the the answer would then be:
To multiply radicals you need like terms, i.e. ∛2∗∛6 then you multiply, and simplify if necessary. In this case:
∛2 ∗ ∛6
another example from the notes is:
∜3⁶w⁷ ∗ ∜3²w²
***Note*** to simplify radicals with variable, if it helps you can
write them out, and box the groups of variables.
/ \ nnnn nnnn nnnn
The answer is 2n³ because the 2's form a group of 4, and the there are 3 groups of n's able to be formed.
Hope this helps, if not you can always look back to the moodle notes or ask Mr. Cope for help.