Question

Someone please explain to me where i went wrong...

Answer 1

Expand and simplify
\[x ^{4}+27x\]
That is the question. this is how I did it.

\[x(x ^{3}+27)\]
\[x(x ^{3}+3^{3})\]
\[x(x+3)(x ^{2}+3^{2})\]
\[x(x+3)(x ^{2}+6x+9)\]
That is my answer, but the right answer is
\[x(x+3)(x ^{2}-6x+9)\]
where did I go wrong?

Answer 2

@Garadon_Shagan

Answer 3

@aryandecoolest @ganeshie8 @hopelovelift @midhun.madhu1987 @waterineyes

Answer 4

Upto \(x(x^3 + 3^3)\) you are \(\checkmark\)..

Answer 5

\[\large x (x^3 + 3^3) \color{red}{\ne} x(x+3)(x^2 + 3^2)\]

Answer 6

You have used here "Your Own Formula" here not "Mathematical" one.. :P

Answer 7

yeah..i have expanded and seen they are not equal...

Answer 8

lol...it looks corrects ....so how do I expand from there

Answer 9

i think answer should be x(x+3)(x^2-3x+9)

Answer 10

Do tell me if I am going directly but the formula for :

\[a^3 - b^3 \; \; is \; \; \large a^3 + b^3 = \color{green}{(a+b)(a^2 + b^2 - ab)}\]

Answer 11

@aryandecoolest yes thats it

Answer 12

yeah that's the formula..!!!

Answer 13

@waterineyes nice!!! let me try and use that

Answer 14

That is for + sign in between..

For - sign in between the formula becomes:

\[\large a^3 - b^3 = \color{blue}{(a-b)(a^2 + b^2 + ab)}\]

Do note the difference between the two formulae.. :)

Answer 15

thnks :D

Answer 16

what about\[3x ^{\frac{ 3 }{ 2 }}-9x ^{\frac{ 1 }{ 2 }}+6x ^{\frac{ -1 }{ 2 }}\]

Answer 17

please take me through

Answer 18

What you want to do with it??

Answer 19

the instructions say Expand and simplify

Answer 20

Can you take firstly 3x^{1/2} common from every term??

Answer 21

will it be
\[3x ^{\frac{ 1 }{ 2 }}(1^{3}-3+2^{-1})\]

Answer 22

Check it once again..

Answer 23

You can write :

\[\large x^\frac{3}{2} = x \cdot x^\frac{1}{2}\]

Answer 24

So tell me one by one, if you take common from first term only, then inside what is left now?

Answer 25

You got that how that came??

Answer 26

ok...will it be
\[3x ^{\frac{ 1 }{ 2 }}(x ^{3}-3+2^{-1})\]

Answer 27

the first x raise to one not 3

Answer 28

is it?

Answer 29

@waterineyes

Answer 30

First time is right...

Answer 31

Second term is also right..

Sorry, my net was slow, so openstudy was not opening up for me..

Answer 32

You need to work out for third term.. :)

Answer 33

\(2^{-1}\) or \(2x^{-1}\) ??

Answer 34

You there??

Answer 35

so 2^-1 is not correct?

Answer 36

Not at all..

Answer 37

See, there is a difference between:

x power half and x power -half, can you tell me??

Answer 38

im stuck :(

Answer 39

-ve you reciprocate?

Answer 40

See, if you having trouble taking common, then make the term like that..

Answer 41

I am only solving for third term as you have done for first two terms right..

Answer 42

yes please

Answer 43

So, for term you have \(6x^{\frac{-1}{2}}\)

Answer 44

Now taking 3 (x raised to power half common):

You firstly do I like this:

You must know if I will multiply something by a quantity say x and then dividing same thing by x, result will be same, Right??

Answer 45

Like for example you have: 2

Now, I multiply 2 by 3 and then divide by 3, then result will be 2 only..

Answer 46

\[2 \implies \frac{2 \times 3}{3} = 2 \; \; only\]

Answer 47

yes..thats right

Answer 48

So, same you can do for third term, and I think you must be knowing by now what I am going to do, as you want to take x power half common, so multiply and divide by this only..

Answer 49

\[\large 6 x^\frac{-1}{2} \implies \frac{6 x^\frac{-1}{2} \cdot \color{red}{x^\frac{1}{2}}}{\color{red}{x^\frac{1}{2}}}\]

Answer 50

huh..im confused here

Answer 51

Am I right there? Are you getting??

Answer 52

@waterineyes let us continue in a few hours, i have to go now...I will let you know when I come back

Answer 53

Okay, I wait here, tell me where are you confused??

Answer 54

Can you tell me exact time when you will come?? I can come online according to that..

Because I am also a Student like you, I have to study on my own like you.. :)

Answer 55

ok...mayb in 3hours time

Answer 56

It is 5:51 in the evening here, So I will come at about 9pm, after 3 hours, 9 minutes, 20 seconds.. Fair enough??

Answer 57

If I could not come, then call Ganesh here, just tag him here, he will definitely help you, don't worry about that.. :)

Answer 58

sure..thanks!

Answer 59

@waterineyes
I was finally able to factor out. this is what i got
\[3x ^{\frac{ 3 }{ 2 }}-9x ^{\frac{ 1 }{ 2 }}+6x ^{\frac{ -1 }{ 2 }}= 3x ^{\frac{ 1 }{ 2 }}(x+2x ^{-1}-3)\]
So what is the next step?

Answer 60

Yes, you are right till here...

Answer 61

Now you can write it as:

\[3 x ^\frac{1}{2} (x + \frac{2}{x} - 3) \implies 3x ^\frac{1}{2}(\frac{x^2 - 3x + 2}{x})\]

Answer 62

Now try to factorize numerator there and :

\[3 \frac{x^\frac{1}{2}}{x} \implies 3 x^{\frac{-1}{2}}\]

So, you get your final answer something like this:

\[3x^{\frac{-1}{2}} (.....) (......)\]

You have to now fill these brackets and you can do so by factorizing numerator there..