Mathematics
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OpenStudy (anonymous):
Please help Quick questions
13 years ago
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OpenStudy (anonymous):
13 years ago
OpenStudy (anonymous):
@iambatman @some_someone @Opcode
13 years ago
OpenStudy (ineedserioushelp):
Good luck getting help on those lol
0_0
13 years ago
OpenStudy (anonymous):
@some_someone can you help me?
13 years ago
OpenStudy (anonymous):
Okay, for #2 do you know what to factor out?
13 years ago
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OpenStudy (anonymous):
no
13 years ago
OpenStudy (anonymous):
the coefficients are: 16, 2, -22
so what is the least common factor of the three of them?
13 years ago
OpenStudy (anonymous):
2
13 years ago
OpenStudy (anonymous):
good
13 years ago
OpenStudy (anonymous):
i think 2 is a am i correct?
13 years ago
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OpenStudy (anonymous):
not quite
13 years ago
OpenStudy (anonymous):
ohh okay
13 years ago
OpenStudy (anonymous):
16a^3b^7 + 2a^6b^4 - 22a^4b^5
so far we have: factor out a 2, but we also need to factor out some of the varaibles
13 years ago
OpenStudy (anonymous):
is it d since you factor some out?
13 years ago
OpenStudy (anonymous):
no not quite
13 years ago
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OpenStudy (anonymous):
16a^3b^7 + 2a^6b^4 - 22a^4b^5
so need to factor out the lowest factor of the variable a and variable b? do you understand me, because it might sound kind of confusing i got to admit.
13 years ago
OpenStudy (anonymous):
so we factor out the lowest degree of the variable.
i.e we factor out a^3 and b^4
so we factor out a 2a^3b^4 out of 16a^3b^7 + 2a^6b^4 - 22a^4b^5
13 years ago
OpenStudy (anonymous):
i guess little it so i know there will be a 2 outside the parenthesis right?
13 years ago
OpenStudy (anonymous):
\[16a^3b^7 + 2a^6b^4 - 22a^4b^5\]
\[2a^3b^4(-----)\]
13 years ago
OpenStudy (anonymous):
okay so its b
13 years ago
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OpenStudy (anonymous):
okay lets do 3 i think its c or a
13 years ago
OpenStudy (anonymous):
yes, but do you know why and how?
13 years ago
OpenStudy (anonymous):
which one?
13 years ago
OpenStudy (anonymous):
for #2
13 years ago
OpenStudy (anonymous):
ohh because of the least common factor for eact
13 years ago
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OpenStudy (anonymous):
yeah okay seem like you understood most of the process for #2.
lets move on to #3 and verify you are understanding (:
13 years ago
OpenStudy (anonymous):
do you know what to do for #3, now that we did #2?
13 years ago
OpenStudy (anonymous):
find the least comon factor for each?
13 years ago
OpenStudy (anonymous):
yes, try it (:
13 years ago
OpenStudy (anonymous):
okay they all go into 9 so its 9x exponents subtract i think
13 years ago
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OpenStudy (anonymous):
does 9 go into 3?
13 years ago
OpenStudy (anonymous):
coefficients: 3, 18, 27
13 years ago
OpenStudy (anonymous):
so what is the least common factor that goes into all three of those coefficients?
13 years ago
OpenStudy (anonymous):
3
13 years ago
OpenStudy (anonymous):
3 nvm
13 years ago
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OpenStudy (anonymous):
good (:
well not 1, because it obviously goes into any number lol, so 3 would be correct.
13 years ago
OpenStudy (anonymous):
now the variables: x^3, x^2, x
13 years ago
OpenStudy (anonymous):
1 so is it c or b @some_someone
13 years ago
OpenStudy (anonymous):
yes its one of those
13 years ago
OpenStudy (anonymous):
but from the variables we would factor out an x
13 years ago
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OpenStudy (anonymous):
so we would factor out a 3x
13 years ago
OpenStudy (anonymous):
3x(______)
13 years ago
OpenStudy (anonymous):
can you do that?
13 years ago
OpenStudy (anonymous):
yes x3+6x3+9x2
13 years ago
OpenStudy (anonymous):
3x^3 + 18x^2 + 27x
factor out a 3x
so:
3x(x^2 + 6x + 9)
your not done yet
u have to factor the thing inside the ( )
13 years ago
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OpenStudy (anonymous):
how?
13 years ago
OpenStudy (anonymous):
have you done any related problems like these? Maybe you have notes or something?
13 years ago
OpenStudy (anonymous):
You seem very lost; sorry to say it.
13 years ago
OpenStudy (anonymous):
no this is like a before quiz pre test
13 years ago
OpenStudy (anonymous):
ouh so u have not been introduced to this material?
13 years ago
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OpenStudy (anonymous):
yes
13 years ago
OpenStudy (anonymous):
so please gve me the answr ans an explenation so i can understand them
13 years ago
OpenStudy (anonymous):
@some_someone is 3 c?
13 years ago
OpenStudy (anonymous):
yes
13 years ago
OpenStudy (anonymous):
okay can you do 6 with an explanation so i can take notes ?
13 years ago
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OpenStudy (anonymous):
what would u factor out? take in mind the 2 other problems we did
13 years ago
OpenStudy (anonymous):
3 the 30 and the 75 they all go into 3
13 years ago
OpenStudy (anonymous):
yes
13 years ago
OpenStudy (anonymous):
what about from the variables?
13 years ago
OpenStudy (anonymous):
take the smallest one with the smallest degree(exponent)
13 years ago
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OpenStudy (anonymous):
is it c
13 years ago
OpenStudy (anonymous):
YES!!! GOOD JOB (:
I feel happy, your understanding a little better now, you are doing good.
13 years ago
OpenStudy (anonymous):
for 7 is it b?
13 years ago
OpenStudy (anonymous):
you could still factor it and reduce more
13 years ago
OpenStudy (anonymous):
a?
13 years ago
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OpenStudy (anonymous):
if there is a - and a + then it would be a -
and there is no negatives
13 years ago
OpenStudy (anonymous):
so c?
13 years ago
OpenStudy (anonymous):
\[5c^5 + 60^4 + 180^3\]
\[5c^3(c^2 + 12c + 36)\]
what two factors multiply to +36 and adds to +12?
13 years ago
OpenStudy (anonymous):
6?
13 years ago
OpenStudy (anonymous):
\[5c^3(c+6)(c+6)\]
\[5c^3(c+6)^2\]
13 years ago
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OpenStudy (anonymous):
yes good job (:
13 years ago
OpenStudy (anonymous):
okay 10 i dont understadn
13 years ago
OpenStudy (anonymous):
lets take a look
13 years ago
OpenStudy (anonymous):
\[d^4 - 8d^2 + 16\]
13 years ago
OpenStudy (anonymous):
we're going to have (....)(....)
13 years ago
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OpenStudy (anonymous):
(d^2....)(d^2....)
do you know why?
13 years ago
OpenStudy (anonymous):
no
13 years ago
OpenStudy (anonymous):
because if you FOIL back, d^2 * d^2 = d^4 which is the 1st term in ur expression
13 years ago
OpenStudy (anonymous):
ohh okay is it a?
13 years ago
OpenStudy (anonymous):
no, (d^2 - 4)(d^2+4) will give you: d^4 + 4d^2 -4d^2 - 16
13 years ago
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OpenStudy (anonymous):
ohh okay so its b
13 years ago
OpenStudy (anonymous):
@some_someone
13 years ago
OpenStudy (anonymous):
yep
13 years ago
OpenStudy (anonymous):
thanks
13 years ago
OpenStudy (anonymous):
no prob
13 years ago