Question

What are all the possible rational zeros for `f(x)=x^3+3x^2+x-2` and how do you find all zeros?

Answer 1

All the zeroes of `f(x)` are,`-2,(-1+-sqrt5)/2;` among these,

only rational zero is `-2`.

Explanation:

We observe that the sum of the co-effs. of

`f(x)" is "1+3+1-2=3ne0, :. (x-1)` is not a factor.

Also, the sum of the co-effs. of odd-powered terms is 2, and that

of even-powered 1, Since, these are not equal, so, `(x+1)` is also

not a factor.

Now, `(px+q) | (x^3+3x^2+x-2) rArr p | 1 and q|( -2).`

Hence, the possible linear factors `(px+q)" of "f(x)` are,

`(+-x+-1), (+-x+-2)`

We have already verified that `(+-x+-1)` are not factors.

For `(x+2), "we find, "f(-2)=-8+12-2-2=0`

`:. (x+2)" is a factor"`.

Now, `f(x)=x^3+3x^2+x-2`

`=ul(x^3+2x^2)+ul(x^2+2x)-ul(x-2)`

`=x^2(x+2)+x(x+2)-1(x+2)`

`=(x+2)(x^2+x-1)`

For the Quadratic `x^2+x-1, Delta=b^2-4ac=5.`

`:." its zeroes are, "(-b+-sqrtDelta)/(2a)=(-1+-sqrt5)/2`.

Thus, all the zeroes of `f(x)` are,`-2,(-1+-sqrt5)/2;` among these,

only rational zero is `-2`.

Enjoy Maths.!

Answer 2

All the possible rational zeros for f(x) are `+1;-1;+2;-2`;
all zeros are -2 (rational) and `(-1+-sqrt5)/2` (not rational)

Explanation:

All the possible rational zeros for f(x) are the factors of the known term 2:

`+1;-1;+2;-2`

You would apply the remainder rule to find the first zero:

`color(red)(f(1)=1+3+1-2=3!=0)`

and you conclude that 1 isn't a zero for f(x);

`color(red)(f(-1)=(-1)^3+3(-1)^2+(-1)-2=-1+3-1-2=-1!=0)`

and you conclude that -1 isn't a zero for f(x);

`color(red)(f(2)=2^3+3*2^2cancel(+2)cancel(-2)=8+12=20!=0)`

and you conclude that 2 isn't a zero for f(x);

`color(green)(f(-2)=(-2)^3+3*(-2)^2+(-2)-2=-8+12-2-2=0)`

then -2 is a zero for f(x)

Then you would divide f(x) by (x+2) and get the polynomial:

`x^2+x-1`

that, solved by the quadratic formula, will give the not rational zeros:

`(-1+-sqrt5)/2`