Question

Can you show the work of this problem?
I'm trying to see an easy way of solving this problem but I know the answer...

Answer 1

what problem

Answer 2

Factor and determine the zeros.
\[x^3+3x^2-x-3\]

Answer 3

guess

Answer 4

since the problems are usually cooked up to be easy, the solution is usually an integer
if it is an integer, it must divide 3, and the only integers that divide 3 evenly are \(\pm1,\pm3\)

Answer 5

\[f(1)=0\]by inspection, because
\[1+3-1-3=0\] so we get it on the first try
that is why i said "guess" because really i mean check the obvious choice first

Answer 6

once you know a zero, you can factor
in this case you will get
\[x^3+3x^2-x-3=(x-1)(something)\] and you can find the something either by long division ( a pain)
synthetic division (very snappy)
thinking (easiest method)

Answer 7

I know what the zeros are: (1,0) (-1,0) (-3,0)
But I need to know the steps of how to get to the zeros...

Answer 8

We have x^2 + 3x^2 - x - 3
So factor out x^2 from first two terms & factor out -1 from third & fourth term
that is x2(x + 3) - 1(x + 3)

now factor out (x+3) there fore (x+3) (x^2 - 1)

now we know that (a^2 - b^2) = (a - b)*(a+b)
So we will get (x+ 3)(x+1)(x-1) Hope this will help u :)

Answer 9

then equate these factor to zero & find zeros

Answer 10

Thanks so much :)