Can someone help me with/teach me my math? I've missed some school and im not sure how to do it.

12x^5-4x^4+7x^3-2x^2-5x+2

Find the zeros of the polynomial as the product of prime binomial factors. Determine all the possible rational roots and the y-intercept. And determine if there are any prime quadratic factors over the real number field.

The y intercept is (0,2) ... Plug in zero for x.

By the rational root theorem:
The possible rational roots are (factors of 2)/( factors of 12)

= +/-{ 1,2}/ { 1,2,3,4,6,12}

Now make all possible ratipnal roots= +/-{ 1,1/2,1/3,1/4,1/6,1/12, 2, 2/3 }

12x^5-4x^4+7x^3-2x^2-5x+2
First try 1 and -1. (Add these points to your graph as you go)

f(1)= 12-4+7-2-5+2= 10

f(-1)= -12-4-7-2+5+2= -18

Since there is a sign change , the function crosses the x axis at least once between -1 and 1.
And since f(0)=2, then we can narrow it down between 0 and -1.

Now the best way to try the other fractions is by synthetic division ( you know how to do this, right?)

-1/2). 12... -4.... 7.... -2... -5...2

------ ----(-6)---(5)---(-6)--(4)---(1/2)
...........12...-10....12.....-8...-1.… 2.5

So f(-1/2)= 2.5.....add this point to your graph.
Notice it is higher than the y intercept.

-2/3) 12....-4.. ....7... ...-2.....-5.....2

-----------(-8)- --(8)---(-10)---(8)---(-2)
.........12....-12....15... -12........3. ...0..< remainder is (0) and this row of numbers is the other factor
F(-2/3)= 0

Add this point to your graph.

This means the f(x) factors as (x+2/3)( 12x^4-12x^3+15x^2-12x+3)
= 3(x+2/3)(4x^4-4x^3+5x^2-4x+1)
=(3x+2)(4x^4-4x^3+ 5x^2-4x+1)

Using rational root thm again, we can try +/- 1/2 or 1/4

1/2) 4..... -4.... 5.... -4.... 1
------------(2)--(-1)--(2)---(-1)-
........4..... -2......4... .(-2)...(0)<----- remainder is zero.

(x-1/2) is a factor , leaving (4x^3-2x^2+4x-2)

==(3x+2)(x-1/2)(4x^3-2x^2+4x-2)

Factor out 2 on the cubic: