Question

How do you find all the zeros of `f(x)=x^3+2x^2-x-2`?

Answer 1

`x=-2, 1, -1`

Explanation:

So, I like to factor this sort of problm using synthetic division. First thing's first, let's set up this problem.

`color(white)(-2)|` `1color(white)(000)2color(white)(000)-1color(white)(000)-2`
`color(white)(-2)|`
`-2|`__________

We bring down the `1`, which gives us this:

`color(white)(-2)|` `1color(white)(000)2color(white)(000)-1color(white)(000)-2`
`color(white)(-2)|`
`-2|`__________
`color(white)(0)color(white)(000)1`

We then multiply the `-2` by the `1` that we brought down to give us `-2`. That value is brought up to the next row, which brings us to this:

`color(white)(-2)|` `1color(white)(000)2color(white)(000)-1color(white)(000)-2`
`color(white)(-2)|` `color(white)(00)-2`
`-2|`__________
`color(white)(0)color(white)(000)1`

From here, we just add the `2` to the `-2` and end up at this:

`color(white)(-2)|` `1color(white)(000)2color(white)(000)-1color(white)(000)-2`
`color(white)(-2)|` `color(white)(00)-2`
`-2|`__________
`color(white)(0)color(white)(000)1color(white)(000)0`

If we continue this system, we end up with this:

`color(white)(-2)|` `1color(white)(000)2color(white)(000)-1color(white)(000)-2`
`color(white)(-2)|` `color(white)(00)-2color(white)(0000)0color(white)(000000)2`
`-2|`________
`color(white)(0)color(white)(000)1color(white)(000)0color(white)(00)-1color(white)(000000)0`

So now we have `x=-2` or `x+2=0`, or `(x+2)` and `(x^2-1)`. This can be simplified to just `(x+1)(x-1)`.

We now have the factors of `x^3+2x^2-x-2`, which are `(x+2)(x+1)(x-1)`, which can be rewritten as: `x=-2,-1,1`.

So let's just double check that we found all the zeros of `x^3+2x^2-x-2` be graphing it

graph{x^3+2x^2-x-2}

And look, the three `x`-intercepts are what we said, `x=1,-1,-2`.

Answer 2

The zeros are `-2, -1, 1.`

Explanation:

Method 1

`f(x)=x^3+2x^2-x-2=x^2(x+2)-1(x+2)=(x+2)(x^2-1)=(x+2)(x+1)(x-1).`

Hence, the zeros are `-2, -1, 1.`

Method 2

Observe that the sum of the co-effs. `=1+2-1-2=0`

So, `(x-1)` is a factor of poly. `f(x)`. Now, we will rearrange the terms of the poly. in such a way that `(x-1)` can be taken out as a common factor. See the following computation :-

`f(x)=x^3+color(red)2x^2-color(blue)1x-2=x^3-color(red)1x^2+color(red)3x^2-color(blue)3x+color(blue)2x-2=x^2(x-1)+3x(x-1)+2(x-1)=(x-1)(x^2+3x+2)=(x-1){x^2+(2+1)x+(2xx1)}=(x-1){x^2+2x+x+2}=(x-1){x(x+2)+1(x+2)}=(x-1)(x+2)(x+1).`

Hence, the zeros are `1, -2, -1` as before!