Essentially, we are in REVIEW MODE for Thursday's UNIT 8 TEST!!
SECTION 8-8 FACTORING BY GROUPING IS NOT ON THE TEST!!
The hw posting on MathChamber is up-to-date through the end of the unit.
Remember... ALWAYS LOOK FOR GCF's first... then, if you have a quadratic, give the "SYMMETRY OF THE BOX" a try. Knowing the SPECIALS will save you time, but the BOX will always work.
Watch for some videos this weekend... I am taking "REQUESTS" for specific problems if you get them to me early enough.
can you go over problem #35, i know that you mention it on the hw page, but can you do a little walktrough?
ReplyDeletethanks
For: r²+19rs+90s²
DeleteMake a diamond problem where 90 is your product and 19 is your sum. You should come out with 10 an 9. Because there are two variables r AND s will be in the parenthesis. If you look back at problem 5 on pg 514 r and s are basically x and y. If you use that situation and apply it to what you have now: 10, 9, r and s. Plug in the #'s as if it were problem 5.
You end up with: (r+9s)(r+10s).
Hope this helped.
Helped me! Couldn't have said it better!
DeleteCan someone help me with problems 13 and 20get on Problem Set A?
ReplyDeleteFor #13a I have 15 and 2 as my factors because I divided 60 and 26 in half, due to the 2r², but the 2 needs to be negative for it to work, yet the 30negative isn't negative. I had the samenegative problems with #20. TY
For ProbSetA #13:
Delete2r^2 + 26r + 60
2(r^2 + 13 + 30)
Box it, and you need two numbers that have a product of 30 and a sum of 13... how about 10 & 3?
2(r+10)(r+3)
We're fully factored!!
For ProbSetA #20:
x^2 + 5x + 6
Box it, and you are looking for two numbers with a product of 6 and a sum of 5... how about 3&2?
(x+3)(x+2)
number 30 in problem set A?
ReplyDeleteMake your diamond problem with -12 as the product and 1 as the sum. 1 is going to be some because it's the coefficient of n that is know, but not shown. -3 and 4 fit into the diamond problem so plug them into your answer. You multiply n minus your negative # by n plus your positive #.
ReplyDeleteYour final answer should be (n-3)(n+4)
1 is going to be the SUM. Sorry for the lack of editing.
ReplyDeletefyi... the period2 blog has a good conversation going... it might pay to take a look...
ReplyDeleteokay thanks
DeleteI need help on #44 in the book on pg. 537 in the chpt. review.
ReplyDelete6k^2 - 10kp + 4p^2
Delete(Note: I changed the letter (variable) l to p to avoid confusion with the number 1)
Well, if this is one of the few you are having trouble with, you're in very good shape... the double variable ones are a bit tougher, you won't see a lot of these on the test.
FIRST, look for a GCF... they are all even coefficients, so 2 is your GCF!
2(3k^2 - 5kp + 2p^2)
NOW, the BOX is the THING!
Box it, and you have 3k^2 in the lower left and 2p^2 in the upper right, yes?
We know that the cross-products are equal, so you need two numbers with a product of 6 and a sum of -5... hmmm... search the available factors of 6 and I think you will find -3 & -2.
The double variables make the boxing a little trickier, but basically you'll fill in the open spots with -3kp and -2kp, yes?
That said, do the GCF "dance" across the rows and down the columns to find your binomial terms and you're done.
2(3k-2p)(k-p)
Capeesh?