# Alevel Maths - Finding the Roots of functions

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Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!

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#2

(Original post by

Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!

**tomas530**)Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!

a(variable)^2 + b(variable) + c

So in your case the (variable) is just x^3

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#3

(Original post by

Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!

Attachment 958262

Attachment 958264

**tomas530**)Hi everyone! please could someone explain to me the the part of the solution working that I drew an arrow to? I really dont understand that bit. thanks!

Attachment 958262

Attachment 958264

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6

Then they factorise the new equation for (k-1)(k+8)=0

Now replace the k with x^3 again and solve the brackets

Last edited by laurawatt; 3 weeks ago

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(Original post by

Hi! What they did there was pull a factor of x^3 out, to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6

Then they factorise the new equation for (k-1)(k+8)=0

Now replace the k with x^3 again and solve the brackets

**laurawatt**)Hi! What they did there was pull a factor of x^3 out, to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6

Then they factorise the new equation for (k-1)(k+8)=0

Now replace the k with x^3 again and solve the brackets

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**laurawatt**)

Hi! What they did there was pull a factor of x^3 out, to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6

Then they factorise the new equation for (k-1)(k+8)=0

Now replace the k with x^3 again and solve the brackets

Find all the roots of the following function:

m(x) = x^3 + 5x^2 -24x

But Here you cant just pull out a factor of x to the power of something. So how should I start? Thanks

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#6

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Hi, Im practising some more of these questions and came across this one:

Find all the roots of the following function:

m(x) = x^3 + 5x^2 -24x

But Here you cant just pull out a factor of x to the power of something. So how should I start? Thanks

**tomas530**)Hi, Im practising some more of these questions and came across this one:

Find all the roots of the following function:

m(x) = x^3 + 5x^2 -24x

But Here you cant just pull out a factor of x to the power of something. So how should I start? Thanks

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**tomas530**)

Hi, Im practising some more of these questions and came across this one:

Find all the roots of the following function:

m(x) = x^3 + 5x^2 -24x

But Here you cant just pull out a factor of x to the power of something. So how should I start? Thanks

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#8

(Original post by

Oh I think ive got it. I just pull out x as a factor, right?

**tomas530**)Oh I think ive got it. I just pull out x as a factor, right?

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#10

**laurawatt**)

Hi! What they did there was pull a factor of x^3 out, to make the factorising easier!

To think about it a different way, for the x^6 + 7x^3 - 8 =0, they in effect replaced any x^3 with “k” (just a constant like “x” is), to get k^2 + 7k - 8 = 0, which you’ll notice is a lot easier to factorise than anything with a power of 6

Then they factorise the new equation for (k-1)(k+8)=0

Now replace the k with x^3 again and solve the brackets

However, in the most recent example, indeed a factor of x needs to be pulled out first.

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#11

(Original post by

Slight care is needed with the wording. Here x^3 is not 'pulled out as a factor' at all.

However, in the most recent example, indeed a factor of x needs to be pulled out first.

**RDKGames**)Slight care is needed with the wording. Here x^3 is not 'pulled out as a factor' at all.

However, in the most recent example, indeed a factor of x needs to be pulled out first.

Last edited by laurawatt; 3 weeks ago

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Alright, I have done a few more correctly, but... here comes a hard one. Could one of you please help me with this new question before I make a frustrating mistake?

The question is:

Find all real roots of the following function:

m(x) = 2x^2/3 + 2x^1/3 -12

Would i need to convert them into surds? Or should I 'factor out' ^1/3 ?

Thanks

PS: sorry for so many questions

The question is:

Find all real roots of the following function:

m(x) = 2x^2/3 + 2x^1/3 -12

Would i need to convert them into surds? Or should I 'factor out' ^1/3 ?

Thanks

PS: sorry for so many questions

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#13

(Original post by

Alright, I have done a few more correctly, but... here comes a hard one. Could one of you please help me with this new question before I make a frustrating mistake?

The question is:

Find all real roots of the following function:

m(x) = 2x^2/3 + 2x^1/3 -12

Would i need to convert them into surds? Or should I 'factor out' ^1/3 ?

Thanks

PS: sorry for so many questions

**tomas530**)Alright, I have done a few more correctly, but... here comes a hard one. Could one of you please help me with this new question before I make a frustrating mistake?

The question is:

Find all real roots of the following function:

m(x) = 2x^2/3 + 2x^1/3 -12

Would i need to convert them into surds? Or should I 'factor out' ^1/3 ?

Thanks

PS: sorry for so many questions

^{1/3}

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#15

(Original post by

how though?

**tomas530**)how though?

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