Maths Help - PLease!!

Cathal93

confused by mock question :

2x-t is a factor of 2x^3 + px +p

show that p= -t^3 (all over) 2(t+2)

grrrrrrrrrr

Cathal93 wrote: »

confused by mock question :

2x-t is a factor of 2x^3 + px +p

show that p= -t^3 (all over) 2(t+2)

Divide in?? since its a factor it should divide in evenly?

as in long division!!

or else you sub in t/2 for x!(as 2x-t=> x=t/2). one of them is right anyway...i hope

H2student

Couldn't do this myself. I think you should warn people in the topic title that this is mock related or post in the topic for the mocks since not everyone have finished their mocks yet.

BL1993

2x-t is a factor. Therefore 2x-t=0....x=t/2

Now, sub this into the actaul equation and you get:

2(t/2)^3+p(t/2)+p which according to the theorm should equal zero.

Therefore, after multiplying t/2 into the equation, we get:

(2t^3/8)+(pt/2)+p=0

Multiply by 8:

2t^3+4pt+8p=0

Divide by 2:

t^3+2pt+4p=0

Now, subtract by t^3....2pt+4p=-t^3
Factorise: p(2t+4)=t^3
Divide by (2t+4)....p=-t^3/(2t+4)....p=-t^3/2(t+2)

Hope that helps.

grrrrrrrrrr

BL1993 wrote: »

2x-t is a factor. Therefore 2x-t=0....x=t/2

Now, sub this into the actaul equation and you get:

2(t/2)^3+p(t/2)+p which according to the theorm should equal zero.

Therefore, after multiplying t/2 into the equation, we get:

(2t^3/8)+(pt/2)+p=0

Multiply by 8:

2t^3+4pt+8p=0

Divide by 2:

t^3+2pt+4p=0

Now, subtract by t^3....2pt+4p=-t^3
Factorise: p(2t+4)=t^3
Divide by (2t+4)....p=-t^3/(2t+4)....p=-t^3/2(t+2)

Hope that helps.

this is what i was saying but didnt go to that effort!! ha

Cathal93

Hey thanks so much! I'm sooo stuck on these papers

On examcraft paper 1 I cant work out

1c
6b
7c
8c

Can anyone help???

Overature

yeah, caus its a factor you can divide it by long division and there will be no remaindoor. then you can put what ever answer you got out of that and rearange it into what ever proof that you need to prove

grrrrrrrrrr

Cathal93 wrote: »

Hey thanks so much! I'm sooo stuck on these papers

On examcraft paper 1 I cant work out

1c
6b
7c
8c

Can anyone help???

is examcraft the one with the tick???

BL1993

If you give me the questions, maybe I could help. Please take note that I am only in fifth year so I would not have the whole course covered. I am sure I can help with the Q.1 c part though.

Oh, I would just like to note that in a question similar to the one I just done, you divide by the factor is the factor is in the form ax^2+bx+c or something with a power in it, if it is in the form ax+b, then do what I did.

Cathal93

Show that z^2 -4z +5 is a factor of z^3 +(i-4)z^2 +(5-4i)z +5i
Hence find the 3 roots of z^3 etc =0

grrrrrrrrrr

Cathal93 wrote: »

Show that z^2 -4z +5 is a factor of z^3 +(i-4)z^2 +(5-4i)z +5i
Hence find the 3 roots of z^3 etc =0

divide z^2 -4z +5 into z^3 +(i-4)z^2 +(5-4i)z +5i and you'll get a constant. if you sub in that constant to z^3..... you should gwt 0... then continue from there.


someone else will prob do it for you but i couldnt really be bother. that'll start you at least!!