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Mathematics 8 Online
OpenStudy (anonymous):

Factor 6x^4+6x^3-12x^2

OpenStudy (anonymous):

well first write everything you see in that phrase

OpenStudy (anonymous):

like 6 , x^4 , x^3 and etc.write all of them.

OpenStudy (anonymous):

extract 6 outside

OpenStudy (anonymous):

@Ankh , DON"T just give answers ! he/she wants to be assisted!

OpenStudy (anonymous):

@Twilliams250 , so we have 6, x^4 , x^3 , x^2 , 12 between 6 and 12, 12 is less than 12.so we write 6: 6 .... and between x^4 , x^2 , x^3 which of them is the least?

OpenStudy (anonymous):

x^2

OpenStudy (anonymous):

6x^4 + 6x^3 -12x^2 (get the biggest number you can think of out of all members) You see the biggest one is 6x^2 if you remove 6x^2 from each member, you get: x^2+x-2 Keep in mind you still have to put 6x^2 somewhere But first let's deal with x^2+x-2 If you solve this using the quadratic formula you get the answers x=1, x=-2 what nullifies those numbers? their opposites, therefore (x+2)(x-1) And remember that 6x^2? you just have to add it now: 6x^2(x+2)(x-1)

OpenStudy (anonymous):

so we write x^2 too: 6x^2 ... and then find this: \[\frac{ 6x ^{4} + 6x ^{3} - 12x ^{2} }{ 6x ^{2} }\]

OpenStudy (anonymous):

so you got your last question? first let's factor 2x +8 between 2 and 8, 2 is the least number so we write 2 ... and then 2 times x + 4 would be 2x + 8 so we write 2(x+4) ok then

OpenStudy (anonymous):

and for xy + 4 we can't factor anything!

OpenStudy (anonymous):

so it would be : 2(x+4)(xy+4)

OpenStudy (anonymous):

got it?

OpenStudy (anonymous):

ok,ask

OpenStudy (anonymous):

let f(x)=1-x+x^2 and g(x)=1+x then f(x)g(x) equals A.1-x+x^2=x^3 B.1-x+x^3 C.1-x^2+x^3 D.1+x^3

OpenStudy (anonymous):

you mean f(x) . g(x) or f(g(x)) ?

OpenStudy (anonymous):

f(x)g(x)

OpenStudy (anonymous):

well, i think it's f(x) . g(x) . f(x) = 1 - x + x^2 g(x) = 1 + x so you only need to find (1 - x + x^2)(1-x) ;)

OpenStudy (anonymous):

and can you do the rest ?

OpenStudy (anonymous):

trying

OpenStudy (anonymous):

@Twilliams250 , found it?

OpenStudy (anonymous):

noo can u help

OpenStudy (anonymous):

sry, i didn't see your comment . yes.first find (1-x+x2)(1)

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i think the answer would be A is that correct

OpenStudy (anonymous):

no,the thing is that none of them is the answer!

OpenStudy (anonymous):

what should i do thats a queston on my quiz im taking

OpenStudy (anonymous):

oh sorry wait a moment! i made a mistake :D :P

OpenStudy (anonymous):

yes i found the answer,it isn't A.try it again.

OpenStudy (anonymous):

o k

OpenStudy (anonymous):

i got c this time

OpenStudy (anonymous):

this is false again :D

OpenStudy (anonymous):

how do u get these answers?

OpenStudy (anonymous):

d?

OpenStudy (anonymous):

yes,this time it's correct :)

OpenStudy (anonymous):

so u got it?

OpenStudy (anonymous):

yea i may have another ?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Let f(x)=x^1999 and g(x)=2x-x^5.Then f(x)G(x) equals A. 2x^2000-x^2004 B.(2x-x^5)^1999 C. 2x1999-x9995 D.2x-x^5+x^1999

OpenStudy (anonymous):

again you should find: (x^1999)(2x - x^5) = ?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

well,what did you find?

OpenStudy (anonymous):

C

OpenStudy (anonymous):

let me check

OpenStudy (anonymous):

no no you're incorrect,check it again.

OpenStudy (anonymous):

D?

OpenStudy (anonymous):

no :D

OpenStudy (anonymous):

ok let me try again

OpenStudy (anonymous):

B!

OpenStudy (anonymous):

\[a(c+b) = ac + ab\]

OpenStudy (anonymous):

NO! A :|

OpenStudy (anonymous):

do u want to know how?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

\[(x ^{1999} ) (2x - x ^{5}) = {(x ^{1999})(2x)} - {(x ^{1999})(x ^{5}) }\]

OpenStudy (anonymous):

@Twilliams250

OpenStudy (anonymous):

Ok i got another ? Which of the following polynomials in standard form is equivalent to the polynomial below (x^2y+y^3)(x+xy^2) A.x^3y+xy^3 B.x^3y+xy^3+x+xy^2 C.x^3y+xy^3+x^3y^3+xy^5 D.x^2y+y^3+x+xy^2

OpenStudy (anonymous):

first do the multiplication...

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i got D

OpenStudy (anonymous):

no D is incorrect :) solve these questions carefully!

OpenStudy (anonymous):

A

OpenStudy (anonymous):

no!

OpenStudy (anonymous):

can u explain

OpenStudy (anonymous):

\[(x ^{1999} ) (2x - x ^{5}) = {(x ^{1999})(2x)} - {(x ^{1999})(x ^{5}) } \] just like it!

OpenStudy (anonymous):

ok let me try

OpenStudy (anonymous):

B i think i got it

OpenStudy (anonymous):

no C!

OpenStudy (anonymous):

ok got 5 more ? on the quiz

OpenStudy (anonymous):

ok,go on.

OpenStudy (anonymous):

@Twilliams250 , where are u ? :D

OpenStudy (anonymous):

give all the 5 questions together.

OpenStudy (anonymous):

is it true that x^a x x^b=x^b x x^a. which property of the of polynomial multplication ensures the equality A. Associative B.commutative C.Distributive

OpenStudy (anonymous):

it's B; we call this rule "commutative"

OpenStudy (anonymous):

let f(x)=11-25x-3x^2+4x^3 and g(x)=-3+11x-25x^2 Then F(x)+G(x) equals A. 8+14x-28x^2 B.8-14x-28x^2+4x^3 C. 8+14X-28x^2+4x^3 D.8-14x-28x^2

OpenStudy (anonymous):

what's your idea?

OpenStudy (anonymous):

@Twilliams250

OpenStudy (anonymous):

@Twilliams250 , WHERE ARE YOU?

OpenStudy (anonymous):

sry @Twilliams250 please hurry up.where have u gone? :D

OpenStudy (anonymous):

is the answer B

OpenStudy (anonymous):

yes ;)

OpenStudy (anonymous):

so you're becoming skillful to solving these sorts of questions ;)

OpenStudy (anonymous):

let f(x) be a fifth degree polynomial and G(x) also be a fifth degree polynomial the degree of f(x)+G(x) must also equal 5 A True B False

OpenStudy (anonymous):

what's your idea?

OpenStudy (anonymous):

what

OpenStudy (anonymous):

what's your idea/what's your answer to this question?

OpenStudy (anonymous):

what do u think?

OpenStudy (anonymous):

i think true

OpenStudy (anonymous):

yes ;) true

OpenStudy (anonymous):

are u shore that the answer ?

OpenStudy (anonymous):

yes it's true

OpenStudy (anonymous):

(1+x)(1+y) A. x+y B.1+x+y+xy C.1+x+y D.1+xy

OpenStudy (anonymous):

what's your idea?

OpenStudy (anonymous):

@Twilliams250 , why are you waiting so much?

OpenStudy (anonymous):

i think a

OpenStudy (anonymous):

how (1+x)(1+y) could be x+y? (1+x)(1+y) = 1(1+y) + x(1+y) = ?

OpenStudy (anonymous):

oh is it B

OpenStudy (anonymous):

yes ;)

OpenStudy (anonymous):

are u shore

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