Alevel Maths - Finding the Roots of functions

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!

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!

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

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

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

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

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

Yep

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

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.

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

let y = x^1/3

how though?

this doesn't look right 😩

what must you do to x^1/3 to make it become x^2/3 ?

Rub out the 1 and write 2 in its place?