not sure what its called sorry

[James Smithson]

I have been away from maths for a few years as my wife got very ill but as shes on the mend i started up my couse again . however I am very confused about the smallest details sometimes.

for example in my book we have a deivitive going from

f'(x) = 6 x^2 -9x-6

= 3(2x+1)(x-2)


its the second = part where i am confused i dont know how they got there or how i do simular with the next one. or what its called . I think it was done to remove the exponent to solve for 0 for stationary points but what is the process called where you change it from
6 x^2 -9x-6 to 3(2x+1)(x-2)

im trying to do it with 4x^3 -8x^2 -4x and i get as far as taking the 4x out and end up with 4x(x^2 -2x^2 -x) ...


thanks in advance for help and appologies if in the wrong forum

 

[Dr.Peterson]

I have been away from maths for a few years as my wife got very ill but as shes on the mend i started up my couse again . however I am very confused about the smallest details sometimes.

for example in my book we have a deivitive going from

f'(x) = 6 x^2 -9x-6

= 3(2x+1)(x-2)


its the second = part where i am confused i dont know how they got there or how i do simular with the next one. or what its called . I think it was done to remove the exponent to solve for 0 for stationary points but what is the process called where you change it from
6 x^2 -9x-6 to 3(2x+1)(x-2)

im trying to do it with 4x^3 -8x^2 -4x and i get as far as taking the 4x out and end up with 4x(x^2 -2x^2 -x) ...


thanks in advance for help and appologies if in the wrong forum

The overall problem is calculus, but your difficulty is indeed a matter of basic algebra.

The process is called factorization of a polynomial, which means rewriting an expression as a product of factors. Search for that, and you should be able to find lots of help.

You have a good start on the second problem; now look up how to factorize a (trinomial) quadratic polynomial.

 

[fresh_42]

I have been away from maths for a few years as my wife got very ill but as shes on the mend i started up my couse again . however I am very confused about the smallest details sometimes.

for example in my book we have a deivitive going from

f'(x) = 6 x^2 -9x-6

= 3(2x+1)(x-2)


its the second = part where i am confused i dont know how they got there or how i do simular with the next one. or what its called . I think it was done to remove the exponent to solve for 0 for stationary points but what is the process called where you change it from
6 x^2 -9x-6 to 3(2x+1)(x-2)

im trying to do it with 4x^3 -8x^2 -4x and i get as far as taking the 4x out and end up with 4x(x^2 -2x^2 -x) ...


thanks in advance for help and appologies if in the wrong forum

The idea is that [imath] x-a [/imath] divides a polynomial [imath] p(x) [/imath] if and only if [imath] p(a)=0. [/imath] This means, if you find a zero [imath] a [/imath] with [imath] p(a)=0 [/imath] then you can perform a long division [imath] p(x)\, : \,(x-a) =q(x) [/imath] where [imath] q(x) [/imath] has a lower degree and you can go on finding a number [imath] b [/imath] such that [imath] q(b)=0 [/imath] and divide [imath] q(x)\, : \,(x-b)= r(x) [/imath] etc.

In your example of [imath] p(x)=6x^2-9x-6 [/imath] we can find [imath] p(2)=0 [/imath] by testing low numbers. The second step would then be the division:
[math]\begin{array}{lll} (6x^2-9x-6)\, : \,(x-2)&=6x\\ -6x\cdot (x-2)&=-6x^2+12x\\ \hline \\ 12x-9x-6=3x-6\\ (3x-6)\, : \,(x-2)&=3\\ -3\cdot (x-2)&=-3x+6\\ \hline\\ 0&\\ \hline \end{array}[/math]so that we arrive at [imath] (6x^2-9x-6)\, : \,(x-2)=6x+3=3(2x+1) [/imath] or [imath] 6x^2-9x-6=3(2x+1)(x-2). [/imath] The division works the same way as with numbers, only that we have polynomials here with an [imath] x [/imath] instead of powers of [imath] 10. [/imath]

 

[lookagain]

im trying to do it with 4x^3 -8x^2 -4x and i get as far as taking the 4x out and end up with 4x(x^2 -2x^2 -x) ...

When the middle term is divided by the x-term, the degree of the x-term drops down by one.
When you divide (4x) by (4x), you get 1. The factorization up to that point should look like this:

\(\displaystyle 4x^3 - 8x^2 - 4x \ = \ 4x(x^2 - 2x - 1)\)

It happens to be the case that the trinomial inside the parentheses does not factor into two
binomials with integer coefficients.

 

[James Smithson]

thank you everyone im doing better at this now