quadratic equation

[nidhi]

let p(x)=xcube+axsquare+bx+c where a,b,c are real numbers. ifp(1)=3 and p(2)=6 then the value of p(5)-p(-2) is

[Hymie]

I trust this homework does not have to be submitted tomorrow.

[Anngel123]

Here's a thought. Listen in class. Study. Don't rely on strangers to do your homework.

[factor-fiction]

Hi Hymie- he/she does seem to have quite a bit of A level Maths homework to do this weekend.
What would the value of p(0) be, nidhi?
Have you tried putting the values x=1 and x=2 into the original cubic equation to tell you the values of b and c. You should then be able to get the values of p(5) and p(-2) easily

[factor-fiction]

Okay, I'll have a go at this, but you should check my figures:
p(1)= 1+a+b+c=3
p(2)= 8 +4a +2b +c =6

rearrange the above to give:
a+b +c =2 (call this equation U)
4a +2b+c=-2 (call this equation V)

Subtract U from V:
3a+b= -4 (call this equation W and we’ll use it later)

Now,
By substitution into p(x)= x³+ax²+bx+c we get:
p(5) =125+25a+5b+c
p(-2) = -8+4a-2b +c
Deduct second equation from the first:
p(5) – p(-2)=133+21a+7b
= 133+7(3a+b)
And we know from equation W that 3a+b=-4
So p(5)- p(-2)= 133-28= 105

[factor-fiction]

Did this help, nidhi or had you already worked this out?

[factor-fiction]

Five urgent questions on Sunday but no responses since.
I have a feeling I may have wasted 15 minutes of my time yesterday working on a couple of these

[sandyRoe]

Factor 30, could you explain this?
http://www.theanswerb.../Question1058902.html

[factor-fiction]

Does this work, sandywroe?

Try (11+8-7) ÷ 2 = 3+3

[factor-fiction]

Hi nidhi- I did see your other thread in which you thanked me but that thread seems to have been removed. In my reply (which disappeared with the thread) I explained how you can post a thankyou or clarify queried points simply by submitting this as an 'answer' to your own question. Hope this helps. Have a go.