Question

Identify all of the rational zeros of the polynomial function f(x)=x^3-3x^2+4x-12

Answer 1

We have \[f(x)=x^3-3x^2+4x-12\] and we can factor this out by grouping, such as \[(x^3-3x^2)+(4x-12)=0\] WE can try pulling out a common multiplier between both of these by factoring out these groups to their simplest form, \[x^2(x-3)+4(x-3)=0\] now that we have "(x-3)" as our common multiplier, we can regroup this function. \[(x^2-4)(x-3)=0\] now we solve for (x^2-4) and (x-3) equalling them to0

Answer 2

I got -3 and -4

Answer 3

not quite, see we have \[(x^2-4)=0\] when we solve for x we're going to take the square root of 4, that means we'll get 2 values of the sqrt of 4.\[x^2= 4\]\[x=\sqrt{4}=\pm2\]

Answer 4

Then we have \[(x-3)=0\] solving for x we take 3 to the other side, so we're adding by 3 to both sides. \[x-3+3=+3\]\[x= +3\]

And that gives us the values of \(x= \pm 2, +3\)

Answer 5

Does it kind of make sense?

Answer 6

I got -1,-3,4

Answer 7

How did you get these values? can you tell me the steps you took?

Answer 8

Let me draw it out, maybe that would help?