# Step support programme assignment 7 help pls It is given that x3 +bx2 +cx+d (x−α)(x−β)(x−γ). By substituting 3 different values of x show that:
αβγ = −d
(1 α)(1 β)(1 γ) = 1 + b + c + d
(1 + α)(1 + β)(1 + γ) = 1 b + c d

Given that
where α, β and γ are integers, show using one of the equations from part (iv) that:
x3 −4x2 −4x+16≡(x−α)(x−β)(x−γ), (1 + α)(1 + β)(1 + γ) = −15 .
What are the 8 possible values of α? Find the possible values of 1 + α first. Write down the value of αβγ. Which of the 8 values of α can you now eliminate?
Now use the remaining equation from part (iv) to bring the possible values of α down to three.

Someone pls help with this question I’m stuck. The hints and partial solutions says that to work out 1 + alpha, you need to work out the factors of 15 but I don’t understand why. Also, I don’t get which of the 8 values of alpha to eliminate. Pls help

Scroll to see replies Original post by Fndjdidisjb
It is given that x3 +bx2 +cx+d (x−α)(x−β)(x−γ). By substituting 3 different values of x show that:
αβγ = −d
(1 α)(1 β)(1 γ) = 1 + b + c + d
(1 + α)(1 + β)(1 + γ) = 1 b + c d

Given that
where α, β and γ are integers, show using one of the equations from part (iv) that:
x3 −4x2 −4x+16≡(x−α)(x−β)(x−γ), (1 + α)(1 + β)(1 + γ) = −15 .
What are the 8 possible values of α? Find the possible values of 1 + α first. Write down the value of αβγ. Which of the 8 values of α can you now eliminate?
Now use the remaining equation from part (iv) to bring the possible values of α down to three.

Someone pls help with this question I’m stuck. The hints and partial solutions says that to work out 1 + alpha, you need to work out the factors of 15 but I don’t understand why. Also, I don’t get which of the 8 values of alpha to eliminate. Pls help

If the roots are integers then so are 1+root, so you have
integer*integer*integer = -15
so you need to look at the factors of -15 as the integers on the left hand side must be the factors of -15.
(edited 2 weeks ago) Original post by mqb2766
If the roots are integers then so are 1+root, so you have
integer*integer*integer = -15
so you need to look at the factors of -15 as the left hand side must be the factors of -15.

For some reason I thought that you can only have 2 factors multiply to give -15. Thx for helping. Also how do u know which ones to eliminate ? Original post by Fndjdidisjb
For some reason I thought that you can only have 2 factors multiply to give -15. Thx for helping. Also how do u know which ones to eliminate ?

Have you worked out their values and alpha*beta*gamma as the question suggests.

Factors are usually listed in pairs, but ...
(edited 2 weeks ago) Original post by mqb2766
Have you worked out their values and alpha,beta.gamma as the question suggests.

Factors are usually listed in pairs, but ...

No I haven’t because the question asks for that later Original post by Fndjdidisjb
No I haven’t because the question asks for that later

It asks for the 8 values of alpha, then alpha.beta.gamma, then asks for which you can eliminate? So what have you got for the first two parts? Original post by mqb2766
It asks for the 8 values of alpha, then alpha.beta.gamma, then asks for which you can eliminate? So what have you got for the first two parts?

Re-read the question. It asks for the 8 values of alpha and then the value of alpha x beta x gamma = -d = -16. The.on it asks into eliminate some values of alpha. Original post by Fndjdidisjb
Re-read the question. It asks for the 8 values of alpha and then the value of alpha x beta x gamma = -d = -16. The.on it asks into eliminate some values of alpha.

Were saying the same thing. What do you get for the 8 potential values? Original post by mqb2766
Were saying the same thing. What do you get for the 8 potential values?

(1 + α) = ±1, ±3, ±5, or ±15 so α = 0, −2, 2, −4, 4, −6, 14, or −16 Original post by Fndjdidisjb
(1 + α) = ±1, ±3, ±5, or ±15 so α = 0, −2, 2, −4, 4, −6, 14, or −16

So which of those correspond to potential integer roots mulitplyling to -16?
(edited 2 weeks ago) Original post by mqb2766
So which of those correspond to potential integer roots mulitplyling to -16?

Do you mean three numbers from this list that multiply to give -16 or any number from this multiplied by any 2 integers to give -16? That’s what I’m confused about.
It can’t be 0 for sure.
(edited 2 weeks ago) Original post by Fndjdidisjb
Do you mean three numbers from this list that multiply to give -16 or any number from this multiplied by any 2 integers to give -16? That’s what I’m confused about.
It can’t be 0 for sure.

Just like the previous part, the three roots are integers and multiply to -16, so alpha, beta and gamma must both be factors of -16 and in that list so it looks like 3 or 4 values can be rejected for alpha.

Then in the next question part you reject 1 or 2 more.
(edited 2 weeks ago) Original post by mqb2766
Just like the previous part, the three roots are integers and multiply to -16, so alpha, beta and gamma must both be factors of -16 and in that list so it looks like 3 or 4 values can be rejected for alpha.

Then in the next question part you reject 1 or 2 more.

Tysm. Ik the question was really simple but I was just stuck on it. Original post by Fndjdidisjb
Tysm. Ik the question was really simple but I was just stuck on it.

Wont say step questions are simple, but the more practice you do the better you get (not surprisingly). Original post by mqb2766
Wont say step questions are simple, but the more practice you do the better you get (not surprisingly).

It's a bit of a weird question TBH. It's not at all hard to spot the roots directly, making the games with alpha+/-1 etc. seem somewhat pointless.

I think the real STEP questions are better at avoiding artificial "make work" (not withstanding the infamous "hence show that 4+7=11" question).
(edited 2 weeks ago) Original post by DFranklin
It's a bit of a weird question TBH. It's not at all hard to spot the roots directly, making the games with alpha+/-1 etc. seem somewhat pointless.

I think the real STEP questions are better at avoiding artificial "make work" (not withstanding the infamous "hence show that 4+7=11" question).

Its one of the warm up questions on the step 1 modules, so more prescriptive than an actual question and as you say the numbers are fairly small/can pretty much be guessed/verified or easily factorised by spotting the x-4 part
(edited 1 week ago) This is also from step support programme question 7 - warms down

I have a very old set of balancing scales.
i) If I can only put the weights in one of the scale pans:
(a) Show that I can choose just three weights to measure all the integer number of ounces from 1 ounce to 7 ounces, and that there is only one such choice.
(b) Find 5 weights which enable me to weigh out all the integer number of ounces from 1 ounce to 31 ounces.
(c) Show that if I have only n weights, I cannot weigh more than 2^n different weights (including zero ounces). Hint: each weight is either in the pan or not in the pan. How can I choose weights in order to measure all the integer number of ounces from 1 ounce to 2^(n) 1 ounces?

I’ve done parts a and b but I’m stuck on part. I don’t fully understand the explanation they’ve given in hints. Somebody plss help!!
(edited 1 week ago) Original post by Fndjdidisjb
This is also from step support programme question 7 - warms down

I have a very old set of balancing scales.
i) If I can only put the weights in one of the scale pans:
(a) Show that I can choose just three weights to measure all the integer number of ounces from 1 ounce to 7 ounces, and that there is only one such choice.
(b) Find 5 weights which enable me to weigh out all the integer number of ounces from 1 ounce to 31 ounces.
(c) Show that if I have only n weights, I cannot weigh more than 2^n different weights (including zero ounces). Hint: each weight is either in the pan or not in the pan. How can I choose weights in order to measure all the integer number of ounces from 1 ounce to 2^(n) 1 ounces?

I’ve done parts a and b but I’m stuck on part. I don’t fully understand the explanation they’ve given in hints. Somebody plss help!!

Its obv a generalisation of the first two parts. What dont you understand about the partial solution? If you have n weights, you have 2^n combinations of them as each weight is either included or not. So you can weigh things from 0..2^n-1 Original post by mqb2766
Its obv a generalisation of the first two parts. What dont you understand about the partial solution? If you have n weights, you have 2^n combinations of them as each weight is either included or not. So you can weigh things from 0..2^n-1

To weigh 0, you can’t choose any weight. Are u allowed to do that?
Also, the partial solutions say that diff combinations can give the same weight but when counting the no of different weights, do you include both combinations or just 1? Original post by Fndjdidisjb
To weigh 0, you can’t choose any weight. Are u allowed to do that?
Also, the partial solutions say that diff combinations can give the same weight but when counting the no of different weights, do you include both combinations or just 1?

Its a balancing scales, so putting zero weights on is valid to weigh 0.

The whole point of getting all 2^n different combinations is that you want to choose your weights carefully so that you dont have different combinations measuring the same weight, as in part a) and b). Rather each combination generates a different weight and every weight from 0..2^n-1 is able to be generated. Choosing weights which are powers of 2 does that as
first weight 1 gives 0,1
second weight 2 gives 0,1,2,3
third weight 4 gives 0,1,2,3,4,5,6,7
....
nth weight 2^(n-1) gives 0,1,2,3,...2^n-1

As the partial solution mentions its effectively just the binary (base 2) representation of an integer using n bits, and each of the weights are the place values 2^0, 2^1, 2^2, 2^3, ..
(edited 1 week ago)