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#1 2008-06-14 19:13:42

leena
Guest

help needed with factoring problems

1. 64p^4 - p(p to the power 4) = 64 p^3 ?? is this correct...not sure

2. 625 - (t-10)^2(to the power 2) = 625 - (t-10) (t-10) = 625 - t^2 - 20t +100 = 525 - t^2 +20t..not sure if this is correct??

3. 16a ^2 - 81b^2 = (4a +9b) (2a - 3b)(2+3)...also not sure

#2 2008-06-15 03:14:39

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,552

Re: help needed with factoring problems

1.   64p^4 - p     ->    p (64p^3 - 1)

2.  625 - (t - 10)^2     ->     625 - (t^2 - 20t + 100)     ->     (625 - 100) - t^2 + 20t      ->      -t^2 + 20t + 525

3.  Nice guess on #3 but it doesn't work that way.   Learn "Difference between 2 squares" to do it.


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#3 2008-06-15 04:39:00

leena
Guest

Re: help needed with factoring problems

Thanks Alot for your help Frank!!!:))

#4 2008-06-15 04:48:03

leena
Guest

Re: help needed with factoring problems

1. 8 a^2 + 18a - 5
= (4a-1) (2a+5)
4a-1=0    2a+5=0
4a=0+1   2a=0-5
4a=1       2a=-5
a= 1/4       a=-5/2


2. 2y^2 +10y - 132
= 2(y^2 +5y - 66)


3. xw - yw + xz - yz
=wx-wy + zx + zy

Are they correct?? Thanks!!

#5 2008-06-15 06:17:58

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: help needed with factoring problems

1 is correct. I think you only needed to factorise it, but the extra working is correct anyway.

2 is right so far but you can factorise it further, to 2(y-6)(y+11).

3 should be wx - wy + xz - yz (Note the second minus sign). In general, it's common to put the variables of each term in alphabetical order.
Also, you can factorise it further. Try making a split between the second and third terms, factorising each pair separately and adding the results.

Also, by writing the 2nd question in your first post as 25² - (t+10)², you can factorise it using the difference of two squares. That's simpler than expanding everything out first.


Why did the vector cross the road?
It wanted to be normal.

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#6 2013-03-25 10:06:07

Jake6.0
Guest

Re: help needed with factoring problems

64p^4-p
p(64p^3-1)                                                                 factor out p
p(4p-1)(16p^2+4p+1)                                                 use difference of Cubes (a^3-b^3) where a=4p and b=1

answer=p(4p-1)(16p^2+4p+1)

that's your answer. Although, if you really want to factor this thing out, use quadratic formula to factor the prime 16p^2+4p+1.

-4+-root(4^2-4*16*1)                  -4+-root(16-64)               -4+-root(-48)                  -4+-(4{i}root(3))              -1+-{i}root(3)
              2*16                                         32                                 32                                   32                              8

so, x=-1+{i}root(3), and     -1-{i}root(3)                            "{i} is imaginary number"
               8                               8

Thus, final factoring becomes              p(4p-1)(p-[-1+{i}root(3)])(p-[-1-{i}root(3)])
                                                                                 [8]                       [8]
                 
                                                                                                               This is if you want COMPLETE factoring, otherwise, use former answer.

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