# STEP - Question Help - Cambridge Maths

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Pliskin

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#1

Hi, I'm working through STEP orientated worksheets (not the actual past papers) in preparation for STEP I/II exams and I've came across a question in which I'm completely stumped and spent hours trying to reach a solution. Any help/solutions would be greatly appreciated, thanks a lot.

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ghostwalker

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#2

(Original post by

...

**Pliskin**)...

Then try and rewrite your expressions involving those forms. That will take care of the first two parts.

If you've not covered roots of polynomials, then:

Spoiler:

Given your polynomial has three roots, write it as the product of three factors, expand, and compare coefficients.

And watch out for the coeff of x^3

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Given your polynomial has three roots, write it as the product of three factors, expand, and compare coefficients.

And watch out for the coeff of x^3

For the final part, is a root of the polynomial, so .... That's only a small hint, but hopefully may be enough.

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brianeverit

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#3

(Original post by

Hi, I'm working through STEP orientated worksheets (not the actual past papers) in preparation for STEP I/II exams and I've came across a question in which I'm completely stumped and spent hours trying to reach a solution. Any help/solutions would be greatly appreciated, thanks a lot.

**Pliskin**)Hi, I'm working through STEP orientated worksheets (not the actual past papers) in preparation for STEP I/II exams and I've came across a question in which I'm completely stumped and spent hours trying to reach a solution. Any help/solutions would be greatly appreciated, thanks a lot.

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Pliskin

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#4

(Original post by

Knowing that are the three roots of the polynomial, what three equations can you write down about the relationships between the roots, based on the coefficients of the polynomial (I'm assuming you've covered roots of polynomials, at least for quadratics).

Then try and rewrite your expressions involving those forms. That will take care of the first two parts.

If you've not covered roots of polynomials, then:

For the final part, is a root of the polynomial, so .... That's only a small hint, but hopefully may be enough.

**ghostwalker**)Knowing that are the three roots of the polynomial, what three equations can you write down about the relationships between the roots, based on the coefficients of the polynomial (I'm assuming you've covered roots of polynomials, at least for quadratics).

Then try and rewrite your expressions involving those forms. That will take care of the first two parts.

If you've not covered roots of polynomials, then:

Spoiler:

Given your polynomial has three roots, write it as the product of three factors, expand, and compare coefficients.

And watch out for the coeff of x^3

Show

Given your polynomial has three roots, write it as the product of three factors, expand, and compare coefficients.

And watch out for the coeff of x^3

For the final part, is a root of the polynomial, so .... That's only a small hint, but hopefully may be enough.

Any help is majorly appreciated, thanks a lot !!!

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Zenarthra

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#5

**Pliskin**)

Hi, I'm working through STEP orientated worksheets (not the actual past papers) in preparation for STEP I/II exams and I've came across a question in which I'm completely stumped and spent hours trying to reach a solution. Any help/solutions would be greatly appreciated, thanks a lot.

Alright i just had a go at the question:

So a1, a2 and a3 are roots.

Therefore: (x-a1)(x-a2)(x-a3) = 5x^3 - x^2 - 2x +3

Multiplying out left side gives:

x^3 - x^2(a1+a2+a3)+(a2a3 + a1a2 + a1a3) - a1a2a3

Since 1/a1 + 1/a2 + 1/a3 can be written as a1a2 + a2a3 + a1a3 / a1a2a3

Then when we let x = 0 on both sides we see a1a2a3 = -3

next equating coefficients of x from both sides

gives a2a3+a1a2+a1a3 = -2

Hence 1/a1 + 1/a2 + 1/a3 = -2/-3 = 2/3

Can you finish the rest?

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ghostwalker

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#6

(Original post by

I managed to get b) by expanding (a1+a2+a3)^2

**Pliskin**)I managed to get b) by expanding (a1+a2+a3)^2

, but I'm struggling with a) and c) a bit still... I'm trying the same method for c) as b); expanding (a1+a2+a3)^2.

Any help is majorly appreciated, thanks a lot !!!

Any help is majorly appreciated, thanks a lot !!!

For c) I'd go with

since is a root of the polynomial, then we know

Similarly for the other two roots.

Now add your three equations.

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KongShou

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#7

Pliskin

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#8

(Original post by

Alright i just had a go at the question:

So a1, a2 and a3 are roots.

Therefore: (x-a1)(x-a2)(x-a3) = 5x^3 - x^2 - 2x +3

Multiplying out left side gives:

x^3 - x^2(a1+a2+a3)+(a2a3 + a1a2 + a1a3) - a1a2a3

Since 1/a1 + 1/a2 + 1/a3 can be written as a1a2 + a2a3 + a1a3 / a1a2a3

Then when we let x = 0 on both sides we see a1a2a3 = -3

next equating coefficients of x from both sides

gives a2a3+a1a2+a1a3 = -2

Hence 1/a1 + 1/a2 + 1/a3 = -2/-3 = 2/3

Can you finish the rest?

**Zenarthra**)Alright i just had a go at the question:

So a1, a2 and a3 are roots.

Therefore: (x-a1)(x-a2)(x-a3) = 5x^3 - x^2 - 2x +3

Multiplying out left side gives:

x^3 - x^2(a1+a2+a3)+(a2a3 + a1a2 + a1a3) - a1a2a3

Since 1/a1 + 1/a2 + 1/a3 can be written as a1a2 + a2a3 + a1a3 / a1a2a3

Then when we let x = 0 on both sides we see a1a2a3 = -3

next equating coefficients of x from both sides

gives a2a3+a1a2+a1a3 = -2

Hence 1/a1 + 1/a2 + 1/a3 = -2/-3 = 2/3

Can you finish the rest?

"For c) I'd go with

since is a root of the polynomial, then we know

Similarly for the other two roots.

Now add your three equations."

But I can't seem to get a solution from that method

Can you perhaps share how you would do it? Thanks a lot.

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davros

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#9

(Original post by

Thanks a lot for the help!! Greatly appreciated. I tried for c), the way that ghostwalker said to do it

"For c) I'd go with

since is a root of the polynomial, then we know

Similarly for the other two roots.

Now add your three equations."

But I can't seem to get a solution from that method

Can you perhaps share how you would do it? Thanks a lot.

**Pliskin**)Thanks a lot for the help!! Greatly appreciated. I tried for c), the way that ghostwalker said to do it

"For c) I'd go with

since is a root of the polynomial, then we know

Similarly for the other two roots.

Now add your three equations."

But I can't seem to get a solution from that method

Can you perhaps share how you would do it? Thanks a lot.

Part (ii) gave you the sums of squares of the alphas and you should know (or be able to work out) the sums of the alphas themselves in terms of the original coefficients.

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Zenarthra

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#10

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#10

**Pliskin**)

Thanks a lot for the help!! Greatly appreciated. I tried for c), the way that ghostwalker said to do it

"For c) I'd go with

since is a root of the polynomial, then we know

Similarly for the other two roots.

Now add your three equations."

But I can't seem to get a solution from that method

Can you perhaps share how you would do it? Thanks a lot.

Inputing the individual roots and add 3 equations together gives;

5(a1^3 + a2^3 + a3^3)-2(a1+a2+a3)-(a1^2+a2^2+a3^2)+9=0

You can get a1+a2+a3 from equating the coefficients but remember to divide by 5.

(a1^3 + a2^3 + a3^3) is what you want, (a1^2+a2^2+a3^2) is what you already found.

Therefore you can solve c)

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Thereisnospoon!

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TeeEm

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Thereisnospoon!

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TeeEm

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#14

(Original post by

the OPs notes

**Thereisnospoon!**)the OPs notes

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WillWalker23

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#15

By using roots of polynomial laws:

alpha + beta + gamma = -b/a

alpha.beta + alpha.gamma + gamma.beta = c/a

alpha.beta.gamma = -d/a

When you get, 1/alpha + 1/beta + 1/gamma, you can merge the fractions to form (alpha.beta + alpha.gamma + beta.gamma)/(alpha.beta.gamma), this will help with the solution.

Use a similar approach for the second and third questions. E.g

Alpha^2 + Beta^2 + Gamma^2 can be thought of as

(Alpha + Beta + Gamma)^2 - 2(alpha.beta + beta.gamma and alpha.gamma), use this idea for the third question as well.

alpha + beta + gamma = -b/a

alpha.beta + alpha.gamma + gamma.beta = c/a

alpha.beta.gamma = -d/a

When you get, 1/alpha + 1/beta + 1/gamma, you can merge the fractions to form (alpha.beta + alpha.gamma + beta.gamma)/(alpha.beta.gamma), this will help with the solution.

Use a similar approach for the second and third questions. E.g

Alpha^2 + Beta^2 + Gamma^2 can be thought of as

(Alpha + Beta + Gamma)^2 - 2(alpha.beta + beta.gamma and alpha.gamma), use this idea for the third question as well.

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