Year 13 Maths Help Thread

If p(1)=0 then (x-1) is a factor of p(x) from the remainder theorem, so we can rewrite p(x) as q(x)(x-1) for some polynomial q(x).
In this case, dividing p(x) by (x-1) gives a remainder of 2, so we need to subtract two in order to get a remainder of 0, therefore p(x)=(x-1)q(x)-2.

I'm not 100% sure on this though...

If p(1)=0 then (x-1) is a factor of p(x) from the remainder theorem, so we can rewrite p(x) as q(x)(x-1) for some polynomial q(x).
In this case, dividing p(x) by (x-1) gives a remainder of 2, so we need to subtract two in order to get a remainder of 0, therefore p(x)=(x-1)q(x)-2.

I'm not 100% sure on this though...

Well firstly we can eliminate options which prevent p(1)=2 and then investigate the remaining ones, you can quickly see that all of them depend on various conditions satisfied by q(1) apart from one which will always be true, so I'd go for that one.

Can't be true. Plugging x=1 into that would leave you with -2 so p(1)=2 is not satisfied.

Ah yeah, didn't notice that. so you would have to add on 2 rather than subtract?

Well going off your attempt it wouldn't make sense because you said "p(1)=0" which is also false so you cannot use the remainder theorem.

How do you do part (c) ? The particular solution is $x = (\frac{1}{2} + \frac{4t}{3})e^{-3t} + \frac{1}{18}sin3t$

Erm duno wtf is goingn on above.
Consider p(x)-2=r(x)
x=1 is solution to r(x) so r(x)=(x-1)q(x)
so p(x)=(x-1)q(x)+2


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init plebs everywhere

An employer finds that if he increase weekly wages of each worker by £30 and employs one worker less,he reduces his weekly wage bill from £8160 to £7810, taking the weekly wage of each worker as x ,find out the weekly wage of each worker

how do I do this?

You have two unknowns, the number of employees (n) and their weekly wage (x). You can form two equations for the before and after wage bills. You can then solve for x.

The first equation would be nx = 8160. The rest is left as an exercise for the reader

If only the question writer knew the difference between less and fewer ..

what do you mean by the bit in bold

I did a load of working out and somehow I've gotten to 3n^2 - 32n - 816 = 0 which seems like hell to factorise so I'm assuming I've done it wrong,:/

Sorry - I'm a (partial) grammar pedant. It's fewer for countable things, not less.

That quadratic doesn't look correct to me.

You have:
nx = 8160 and (n-1)(x+30) = 7810

Expanding the second gives: (n-1)x + 30(n-1) = 7810

To make the numbers simpler, I then subtracted the second from the first, giving:
x - 30(n-1) = 350

We're after x, so substitute for n (from nx=8160), and you should get an equation that you can solve with the quadratic formula, taking the positive root.

when I do the bit in bold it gives me
x^2 - 320 x + 244800 = 0

that almost looks worse than the last quadratic

That's correct (at least according to my calculations). Calculate the factors.

I got x as 680 so is that the weekly wage of each worker before or after ?

Look at the original equations.

There are 120 * 35 marks awarded. Let x be the number of successful students. Can you write a formula for the total number of marks in terms of the averages of the successful and unsuccessful students?