f(x) = 3x^3+11x^2-67x+21

What are the 3 real zeros of this polynomial ??

Question

f(x) = 3x^3+11x^2-67x+21 

What are the 3 real zeros of this polynomial ??

myininaya · Answer

try find the rational zeros by
finding the factors of 21

myininaya · Answer

$$f(1)=3+11-67+21=14+21-67=35-67 \neq 0$$
so try -1,3,-3,-7,7,-21,21

myininaya · Answer

you are wanting to see if one these will give you f=0

anonymous · Answer

i have 3 and -7 that equal zero but i am having trouble finding the 3rd zero

anonymous · Answer

the last one is a fraction? i think

myininaya · Answer

oh yeah forgot to include some other factors like  factors of 21/factors of 3
but that is good that you found two zeros

myininaya · Answer

we only need one of those zeros to write this as a factor(quadratic)=0

myininaya · Answer

3|  3        11      -67       21
  |              9        60      -21
    ================
      3        20        -7    |  0

myininaya · Answer

$$3x^2+20x-7=0$$

myininaya · Answer

$$3x^2+21x-1x-7=0$$
$$3x(x+7)-1(x+7)=0$$

myininaya · Answer

$$(x+7)(3x-1)=0$$

myininaya · Answer

any questions?

anonymous · Answer

the last complex zero of this polynomial is 1/3 thanks :)

myininaya · Answer

that is right! :)

anonymous · Answer

How do i use the complex zeros for factor F ?

myininaya · Answer

you want to factor the polynomial?

myininaya · Answer

$$f(x)=(x+7)(3x-1)(x-3)$$

myininaya · Answer

if x=a,b,c are the zeros of f, then
f=(x-a)(x-b)(x-c)

anonymous · Answer

i got 3(x-1/3)(x+7)(x-3) thanks :)

myininaya · Answer

yes that is fine you can write it like that :)