Answer:
[tex]\boxed{-3(x+3)(x^2-6x+10)}[/tex]
Step-by-step explanation:
Hello,
As the polynomial has only real coefficients, it means that 3-i is a zero too, because we apply the Conjugate Zeros Theorem.
It means that we can write the expression as below, k being a real number that we will have to identify.
[tex]k(x+3)(x-3-i)(x-3+i)=k(x+3)((x-3)^2-i^2)\\\\=k(x+3)(x^2-6x+9+1)\\\\=k(x+3)(x^2-6x+10)[/tex]
And for x = 0, y = -90 so we can write
-90=k*3*10, meaning that k=-3
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
