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

2 Answers
Jan 9, 2017

Answer:

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.!

Jan 9, 2017

Answer:

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#