Question

which of the following is a zero for the function f(x)=(x-15)(x+1)(x-10)?

Answer 1

If\[a\cdot b\cdot c=0\]then at least one of a,b,c =0.
So what numbers could make one of your three terms zero?

Answer 2

-15 i think @Xeph

Answer 3

\[f(-15)=(-15-15)(15+1)(15-10)\]\[=(-30)(16)(5)\]It didn't go to 0. What went wrong?

Answer 4

it should have been positive

Answer 5

Ignore the second and 3rd terms, I did +15 instead of -15

Answer 6

Yeah, that's right. If you put +15 in, then the first term is +15-15=0

Answer 7

oh s i was right?

Answer 8

thnk you

Answer 9

And when multiplied by the other 2 terms, everything goes to zero. There are 3 solutions though, can you find the other 2?

Answer 10

@Xeph
I have a question, so this function:
f(x)=(x-15)(x+1)(x-10)
Would the other two terms be -1 and 10?
There is a pattern right? Just find the opposite of that function -15, 1, -10 values.
So the three zeros of this function would write out to be:
15, -1, 10?

Answer 11

That's exactly right. In my original reply I said\[a\cdot b\cdot c=0\] Then at least one of a,b,c must be zero. In a function like this it's pretty easy to see x, but

Answer 12

in more complicated examples you may have to work for it, like if you have\[(2x-4)x=0\] then you can see one solution of x is 0, but to find the second solution in (2x-4) you might need to solve the equation\[2x-4=0\]

Answer 13

Wait, how would we find the zeros of that?
The only possible thing I see is:
\[2*2-4=0\]
Where \(x = 2\)?

Answer 14

That's right, in the equation\[(2x-4)x=0\]\[x=0,2\]but did you solve it or was it trial and error?

Answer 15

I do not see what you mean by solve it...
I factored it out if that is what you mean:
\[2x - 4=0\]
\[2 (x-2) = 0\]
I though okay, what makes zero for -2 and thought, oh it is positive 2.
\[2 (2-2) = 0\]
\[2 (0) = 0\]
\[0 = 0\]

Answer 16

Yep that's a method of solution and works perfectly fine. Keep in mind though that if you always hop to the method you last learned, sometimes you miss easier solutions. I did it this way:\[2x-4=0\]\[2x=4\]\[x=\frac{4}{2}=2\]

Answer 17

Factoring was unnecessary here, but worked completely fine

Answer 18

Well I just learned how to factor polynomials, I guess it's a habit that carried over.
Anyway, thanks :-).

Answer 19

You're welcome. For what it's worth, I think both our approaches were good and simple enough to do in our heads. In a way, if you see the common factor 2 in both terms, you can kind of mentally say "well that 2 isn't really relevant" and just look at x-2=0

I just had a problem the other day where I did things in a huge complicated way, got the correct solution with almost a full page of work, then someone pointed out an easy 2-step solution that I missed because I went straight to what I had learned last :)