y=x^3-3x^2-6x+8 what are the apparent zeros of the function graphed above

Question

dumbcow · Answer

http://www.wolframalpha.com/input/?i=plot+x%5E3+-3x%5E2+-6x%2B8

anonymous · Answer

When I have trouble factoring by grouping, I use polynomial long division. It's easy. It's not much different from regular long division. You need to start with one root. So I use trial and error and sub in x = 1 or x = -1 to see which leads the function to be zero. Sometimes I have to try higher numbers. In this case, x = 1 works. 

Putting in a value of x = 1 turns the equation to zero. 
x = 1 means x-1 = 0 so I will divide my function by (x-1) using polynomial long division. It works out well with no remainder. 
The function simplifies to (x-1)(x^2 - 2x - 8). But we're not done. 
The quadratic can factor to give you (x-1)(x+4)(x-2). So the zeros occur at x = 1, x = -4 and x = 2. Double check it by graphing it on a computer.

anonymous · Answer

@dumbcow   what are the apparent zeros in y=x^3-3x^2-6x+8

anonymous · Answer

Thanks

dumbcow · Answer

look at graph to get your guess then verify with long division
all roots are integers here so graph is sufficient

anonymous · Answer

what about
what is a polynomial function in standard form with zeroes 1,2,-3,and -1?

anonymous · Answer

@jam333 @dumbcow

dumbcow · Answer

any given zero "c" can be written as a factor of the polynomial (x-c)

--> (x-1)(x-2)(x+3)(x+1)
multiply it out to get standard form

anonymous · Answer

Standard form means 'highest degree terms first).  x^2 is degree 2.  x^3 is degree 3, etc.  ...
Multiply it out in steps.  (x-1)(x-2) = x^2 - 2x -1x + 2 = x^2 - 3x +2
(x+3)(x+1) = x^2 + 4x + 3
Now multiply these two quadratic equations.  Do you know how?
 (x^2 - 3x + 2)(x^2 + 4x + 3)  (first times first, first times second, etc...)
=  x^4 + 4x^3 + 3x^2 - 3x^3 - 12x^2 - 9x + 2x^2 + 8x +6
Now gather like terms and simplify to get:
x^4 + x^3 - 7x^2 - x + 6 which is a fourth degree polynomial
Check my adding to make sure you get the same answer.

anonymous · Answer

@jam333 @dumbcow 
the choices are 
x^4+2x^3+7x^2-8x+12
x^4+2x^3-7x^2-8x+12
x^4+2x^3-7x^2+8x+12
X^4+2x^3+7x^2+8x+12