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Please Help, or Teach me. i miss this lesson today and this was the given assignment. 1.) 2a+b a+4b a-7b ___________ - ____________ - ____________ a^2-3ab+2b^2 a^2-4ab+3b^2 a^2-5ab+6a^2 2.) 6m 2n ____________ - ___________ m^2+5m+6 m^2+6m+9 3.) m+2 m _____ - _____ 3m+15 5m+25 Thanks in Advance

Mathematics
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r u familiar with factorising?
yes i am
good:) can u factorise a^2-3ab+2b^2?

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Other answers:

is it 2ab^2?
Nope:( Can u tell me two factors of 2 such that their sum is -3?
(x-2)(x-1)?
a^2-3ab+2b^2 the factors of 2 whose sum is -3 are -1 nd -2. split the middle term -3ab using this. then u will get a^2-2ab-ab+2b^2 =a(a-2b)-b(a-2b) =(a-2b)(a-b) Is it clear?
so the factor of the a^2-4ab+3b^2 is -3 -1 also a^2-3ab-ab+3b^2 =a(a-3b)-b(a-3b) =(a-3b)(a-b)?
a^2-5ab+6a^2 = (a-2)(a-3) i dont know next
can u help me with any number there, so that i can answer the others by myself.
good:) so a^2-3ab+2b^2=(a-2b)(a-b) a^2-4ab+3b^2=(a-3b)(a-b) a^2-5ab+6a^2 = (a-2)(a-3)
can u find thw l.c.m of (a-2b)(a-b), (a-3b)(a-b) and (a-2)(a-3)?
i dont know how.
may i ask if this is correct a^2-5ab+6a^2 = (a-2)(a-3) because i didnt split it yet
nope:( sorry i didnt notice it. if it is a^2-5ab+6b^2, then a^2-5ab+6b^2= (a-2b)(a-3b)
so is it all right if a^2-3ab+2b^2=(a-1b)(a-2b) a^2-4ab+3b^2=(a-1b)(a-3b) ?
yup:)
ajprincess, can you solve any number there then ill do the rest so that i can follow. because its 2am here and i need to sleep thanks for your help btw
is it a^2-5ab+6a^2? or a^2-5ab+6b^2?
its written a^2-5ab+6a^2? but i think it b
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Thank you very much ajprincess
aj princess may i ask last one question? where did you start cancelling?
cancel what?
so you didnt cancel anything?
First I factorised the denominators. then i found lcm.
after finding the lcm i multiplied each fraction both up and down to make them have the common denominator. the common denominator is (m+3)^2(m+2) but the denominator of the first fraction is (m+3)(m+2). To make it (m+3)^2(m+2) i multiplied by (m+3). if i multiplied the denominator then i should multiply the numerator by the same in order to keep the fraction the same. getiing it?
Really sorry. have to leave nw.:(
yes, i reviewed again the answer you've given and it canceling anything isnt neccessary
Thank you very much again ajprincess.
Welcome:)
ok then next time you'll find me in a different name because i cant verify any account. i dont know why :)

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