product of the zeroes

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wendywoo

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the product of the zeroes of f(x)= x3+3x2+3x+2 is......??

so my first step was to pull the x out and i got x(x2+3x+2). then i'm not sure what the next step is.
 
wendywoo said:
the product of the zeroes of f(x)= x3+3x2+3x+2 is......??

so my first step was to pull the x out and i got x(x2+3x+2). then i'm not sure what the next step is.
Click to expand...

Say that a,b,c are zeroes of this function

then P(x)=(x-a)*(x-b)*(x-c)

Right ?

now do the right side

x**3-c*x**2-b*x**2+bcx-a*x**2+a*c*x+a*b*x-abc = P(x)

now set x=0

- abc = P(0) right ?

so -P(0) =

( -2 ) is the answer
 
i have no idea what you just did there. so if i don't factor this equation, how can i find the zeros? can you show me your steps in a simpler manner??
 
wendywoo said:
i have no idea what you just did there. so if i don't factor this equation, how can i find the zeros? can you show me your steps in a simpler manner??
Click to expand...
\(\displaystyle x^3+3x^2+3x+2=(x+2)(x^2+x+1)\)
 
wendywoo said:
the product of the zeroes of f(x)= x3+3x2+3x+2 is......??

so my first step was to pull the x out and i got x(x2+3x+2). then i'm not sure what the next step is.
Click to expand...

There are 3 zeros for this function. You understand that.

Let us assume that those zeros are a, b and c

then from the definition of zeros (roots)

p(x) = (x-a)(x-b)(x-c) and p(x) given to be equal to x3+3*x2+3*x+2


(x-a)(x-b)(x-c) = x3+3*x2+3*x+2

at x = 0

(0-a)(0-b)(0-c) = 03+3*02+3*0+2

-a*b*c = 2

a*b*c = -2

Is that better ......
 
way better now thank you! i just didn't know to use the definition of p(x) before. but how do you know if the answer is -2 or 2? what determines that?
 
wendywoo said:
my question is how do you know to factor out (x+2)??
Click to expand...

I bet someone can provide a more "correct" answer, but when I look for factors I look for three criteria.

1) In the original, there are no minus signs, so that tells us there won't be any in the factors either.
2) What are the factors of the "+2" in the original equation? +2 and +1.
3) Of the available factors, do any of them have a sum equal to 3 ("3x")? 2+1=3

So in this case, these three criteria suggest that we should try to factor out either (x+1) or (x+2).
 
wendywoo said:
my question is how do you know to factor out (x+2)??
Click to expand...

If you haphazardly give values to x, at x= -2 you will get a zero for the Polynomial P(x)

that translates to that a root is -2

so (x+2) is a factor

divide P(x)/(x+2)

and you will obtain P(x)=(x+2)*( x**2 + x + 1)
 
wendywoo said:
my question is how do you know to factor out (x+2)??
Click to expand...

To answer your original question - product of three roots - you do not need to factor with "known" roots.

In the previous posts, a, b and c were assumed to be factors, with unknown values. We did not need to know the exact values of the roots - we need to figure out the product only.

However, you can do that for the sake of practice.
 
wendywoo said:
my question is how do you know to factor out (x+2)?
Click to expand...

If you want to factor the polynomial, you could determine that 2 is a root (and, hence, x+2 is a factor) by using the Rational Roots Theorem. Something called "synthetic division" is also helpful, for checking root candidates coming out of that theorem.
 
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