HELP maths question

[destain]

h^3 - 8
h-2

Youre supposed to rationalise the numerator so you can cancel top and bottom..
Probably really stupid and easy but how, I know the answer but I don't get how you would look at it with one glance and figure it out

[kamil9876]

you know that is a zero of so we have a factorization , find the polynomial using your favourite technique.

[Phy124]

Sorry about doing it this way, but latex is down so:
        ______________
h - 2 ) h³ + 0h² + 0h -8|       1h²(h - 2)
      (-) h³ + 2h²             |       2h(h - 2)
                   2h² + 0h    |       4(h - 2)
               (-) 2h² - 4h    |
                           4h - 8 |
                       (-) 4h - 8 |


Therefore; h³ - 8    =    h² + 2h + 4
                   h - 2

Beaten by kamil ah well

[destain]

ahhh i see you divide it using that way
thanks

[Stick]

h^3 - 8
h-2

Youre supposed to rationalise the numerator so you can cancel top and bottom..
Probably really stupid and easy but how, I know the answer but I don't get how you would look at it with one glance and figure it out

This question is actually much simpler than it appears. You'll need to know the difference of perfect cubes rule:









Hope this helps.

EDIT: LaTeX appears to be broken. Rule: a^3-b^3 = (a-b)(a^2+ab+b^2)

h^3-8 / h-2
= h^3-(2)^3 / h-2
= (h-2)(h^2+2h+4) / h-2
= h^2+2h+4

[luffy]

h^3 - 8
h-2

Youre supposed to rationalise the numerator so you can cancel top and bottom..
Probably really stupid and easy but how, I know the answer but I don't get how you would look at it with one glance and figure it out

I know the others answered the question, but if you're looking for that "one glance" answer, then Stick's method is definitely the way to go. If you practice your difference/addition of cubes formulas, then it will take you no more than about 5 seconds.

[nina_rox]

Hey,
Could any tell me how to sketch 2x / modulus of x? Any help is much appreciated!

[luffy]

Hey,
Could any tell me how to sketch 2x / modulus of x? Any help is much appreciated!

Well, when x>0, |x| = x.  --> E.g. if x = 2, |2| = 2 or if x = 3, |3| = 3.
Therefore, for x>0, the graph will be 2x/x = 2.

When x<0, |x| = -x. --> Eg.  if x = -3, |-3| = 3 = -x or If x = -4, |-4| = 4 = -x.
Therefore, for x<0, the graph will be 2x/(-x) = -2.

So, it is essentially a hybrid function where y = -2 for x<0 and y = 2 for x>0 and we have discontinuity points at x = 0 (because you can't divide by 0).

Hope I helped.

[nina_rox]

Hey,
Could any tell me how to sketch 2x / modulus of x? Any help is much appreciated!

Hmm that sort of makes sense but I would have never thought of it though. I started off separating it into two functions but that didn't work... I think I understand it now thank you very much!
Well, when x>0, |x| = x.  --> E.g. if x = 2, |2| = 2 or if x = 3, |3| = 3.
Therefore, for x>0, the graph will be 2x/x = 2.

When x<0, |x| = -x. --> Eg.  if x = -3, |-3| = 3 = -x or If x = -4, |-4| = 4 = -x.
Therefore, for x<0, the graph will be 2x/(-x) = -2.

So, it is essentially a hybrid function where y = -2 for x<0 and y = 2 for x>0 and we have discontinuity points at x = 0 (because you can't divide by 0).

Hope I helped.

Thanks very much!!

[nina_rox]

Hey, I don't know if this is a stupid question or not but when faced with either a hyperbola or a truncus and you need to find domain and range when do you use R/{x} and when do you use (0,infinity)? Is it whenever the graph is all in the positive side say of the y axis then the range would be (0, infinity), if that makes sense? Thank you!

[Stick]

Use R/{x} when there's only a a couple (maximum) points that are exceptions. (0, infinity) and variations should be used if all values beyond a particular point work.

[nina_rox]

Use R/{x} when there's only a a couple (maximum) points that are exceptions. (0, infinity) and variations should be used if all values beyond a particular point work.

oh okay, that sort of makes sense. So for example if the graph was like 1/(x+2) would I say domain is R/{-2} and range is (0,infinity)?

[Phy124]

Use R/{x} when there's only a a couple (maximum) points that are exceptions. (0, infinity) and variations should be used if all values beyond a particular point work.

oh okay, that sort of makes sense. So for example if the graph was like 1/(x+2) would I say domain is R/{-2} and range is (0,infinity)?

Think about a hyperbola - 1/x - and it's domain and range. The graph has not been translated up or down so it will have the same range as the original - R\{0} - and yes the domain will be R\{-2} as it has been translated two units left.

Alternatively, if the graph was a truncus - 1/x² - then the range would be (0, infinty) and hence the graph 1/(x+2)² would also be (0, infinity)

[nina_rox]

ahh.. thank you very much!