It's pretty much simple. Since we can factor a polynomial by its zeros, lets write one of degree nine :
X(X-1)(X-2)(X-3)(X-4)(X-5)(X+1)(X+2)(X+3)= X^9-9X^8+6X^7+126X^6-231X^5-441X^4+944X^3+324X^2-720X
This polynomial is of degree 9 and has exactly 5 strictly positive zeros : 1, 2, 3, 4, 5
And it has 3 negative zeros : - 1, -1, - 3
And it has 0 as a zero too.
There is also this one :
(X-1)(X-2)(X-3)(X-4)(X²+1)(X+1)(X+2)(X+3) = X^9-4X^8-13X^7+52X^6+35X^5-140X^4+13X^3-52X^2-36X+144
It has 4 positive zeros : 1, 2, 3, 4.
It has complex zeros : i and - i
3 negative zeros : - 1, - 2 , - 3
Good Luck
It would be C) 5y = -4x -15
Answer:
no sé
Step-by-step explanation:
jajajajajajajajjajaa
You start by dividing both sides by 3 and you get your answer for x.
