# Factorising Problem!

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#1
hi, for some reason i am having trouble on factorising:

30x - 3x^2 - 72

i am having trouble with signs etc all fitting in.
help really needed. cheers. **
0
16 years ago
#2
(Original post by alex_thompson)
hi, for some reason i am having trouble on factorising:

30x - 3x^2 - 72

i am having trouble with signs etc all fitting in.
help really needed. cheers. **

Can you not just use the formula? Or does it need to be in brackets? Only I can't seem to make it work either...
0
#3
(Original post by alex_thompson)
hi, for some reason i am having trouble on factorising:

30x - 3x^2 - 72

i am having trouble with signs etc all fitting in.
help really needed. cheers. **
i know, i am having the same problems!
0
16 years ago
#4
Is it equal to 0? if it is then all sings reverse to get
3x^2 -30x +72 = 0

which factorises to

(3x - 18) (x - 4) = 0
0
16 years ago
#5
(-x+4)(3x-18) works, expand to test and see how

a useful site when i was doing maths last year is

http://www.mathsnet.net/index.html

click on AS/A2 (might have gcse stuff too)
0
#6
Can you not just use the formula? Or does it need to be in brackets? Only I can't seem to make it work either...
well no.

i am doing differentiation, and is about finding turning points etc for a given equation of a curve.

u need to set it to zero:

0 = 30x - 3x^2 - 72

and i thought that i factorised the thing, then i get two solutions which give me the two different x-coordinates of the turning points, and from then on i can find the y-values of these 2 turnin points and so on....

but i cant factorise it! signs!
0
16 years ago
#7
Can you not just use the formula? Or does it need to be in brackets? Only I can't seem to make it work either...

No wait got it I think!

(3x - 18)(-x + 4)

That right?
0
16 years ago
#8
No wait got it I think!

(3x - 18)(-x + 4)

That right?
Seems right. I can't really be bothered about checking, though.
0
16 years ago
#9
(Original post by nero076)
(-x+4)(3x-18) works, expand to test and see how

a useful site when i was doing maths last year is

http://www.mathsnet.net/index.html

click on AS/A2 (might have gcse stuff too)

lol drat, should have checked before I posted! At least I got it anyway 0
16 years ago
#10
(Original post by alex_thompson)
well no.

i am doing differentiation, and is about finding turning points etc for a given equation of a curve.

u need to set it to zero:

0 = 30x - 3x^2 - 72

and i thought that i factorised the thing, then i get two solutions which give me the two different x-coordinates of the turning points, and from then on i can find the y-values of these 2 turnin points and so on....

but i cant factorise it! signs!
but if you use the formula, get two solutions which you can use as normal...
0
#11
No wait got it I think!

(3x - 18)(-x + 4)

That right?
thanks very much.

yep, that is correct.

thanks all 4 helping.
0
#12
(Original post by kikzen)
but if you use the formula, get two solutions which you can use as normal...
yeahh i know, but i was wanting to try and factorise it instead, bcoz it did factorise.

in an exam, i wud hav just ued the quad. formula.......
0
16 years ago
#13
I dont think you have to factorise, as far as i can remember the turning points are when dy/dx = 0 (after differentiation). So if you differntiate the equation you get 0= 30 -6x so the turning point is when x = 5???
0
16 years ago
#14
(Original post by George-W-Duck)
I dont think you have to factorise, as far as i can remember the turning points are when dy/dx = 0 (after differentiation). So if you differntiate the equation you get 0= 30 -6x so the turning point is when x = 5???
why would you want to find the minima?
0
#15
(Original post by George-W-Duck)
I dont think you have to factorise, as far as i can remember the turning points are when dy/dx = 0 (after differentiation). So if you differntiate the equation you get 0= 30 -6x so the turning point is when x = 5???

original equation: y = 50 - 72x + 15x^2 - x^3

dy/dx = -72 + 30x - 3x^2

gradient = 0 at turning points, therefore):

0 = (3x - 18) (-x + 4)

and so on...........
0
16 years ago
#16
I thought he wanted to find the turning points? the gradient is 0 for turning points
0
#17
(Original post by George-W-Duck)
I thought he wanted to find the turning points? the gradient is 0 for turning points
The question asked me to find the turning points (coordinates of turning points of course) of curve with equation:
y = 50 - 72x + 15x^2 - x^3 and to determine whether these were maximum or minimum turning points.

dy/dx = -72 + 30x - 3x^2

at turning points, gradient must = 0, therefore:

0 = (3x - 18) (-x + 4)

therefore the two turning points on the curve with the given equation are when x = 6 and when x = 4.

when x = 6, y = 50 - (72*6) + (15*6^2) - (6^3)
therefore: when x = 6, y = -58.

when x = 4, y = 50 - (72*4) + (15 *4^2) - (4^3)
therefore, when x = 4, y = -62.

in order to determine whether the points are a max or min. turning point, we differenciate the gradient function to get:

(d^2y)/(dx^2) = 30 - 6x

sub x=6 into above: (d^2y)/(dx^2) = 30 - (6*6) = -6

THEREFORE THE CURVE HAS ONE TURNING POINT AT (6, -58).
THIS IS A MAXIMUM TURNING POINT. [AS (d^2y)/(dx^2) gave a negative solution when x = 6 was substituted into the above equation)

sub x = 4 into above: (d^2y)/(dx^2) = 30 - (6*4) = 6

THEREFORE THE CURVE HAS ANOTHER TURNING POINT AT (4, -62).
THIS IS A MINIMUM TURNING POINT. [AS (d^2y)/(dx^2) gave a positive solution when x = 6 was substituted into the above equation).
0
#18
isnt mathematics fab????

maths rocks (if you can do it)
0
#19
who else in here enjoys maths?
0
16 years ago
#20
Now i understand, your answer is right, i didnt know you had already differentiated, i just thought you had missed that step! sorry! As for maths been great, if you are one of those people that enjoy scratching their heads then yeah it is 0
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