determine the zeros of the polynomial function : f(x) = x^4 - 13x^2 +36

Question

ksaimouli · Answer

factor it set =0

anonymous · Answer

=0

anonymous · Answer

x^4 - 13x^2 + 36 = 0

anonymous · Answer

then factor

anonymous · Answer

http://www.purplemath.com/modules/factquad.htm

anonymous · Answer

Set x^2 = t, and now you have a quadratic in t which you can solve by the factoring method.  Once you have your t's, convert back to x.

campbell_st · Answer

ok just look at the problem this way

$$(x^2)^2 - 13(x^2) + 36 = 0$$

think of the x^2 as a single value

then factorise

$$(x^2 - 4)(x^2 - 9) = 0$$

now solve for x

anonymous · Answer

(x-3)(X+3)(x-2)(x+2) ?

campbell_st · Answer

thats correct so you will have 4 solutions

anonymous · Answer

cant you just write it like (x^2-4)(x^2-9)? < in factored form

anonymous · Answer

It is not completely factored that way.  If you look at each factor it still has an x^2 in it so you have to go further for the 4 solutions.

anonymous · Answer

okay makes sense, thanks!

campbell_st · Answer

no as you need to find the zeros, this can only be done by solving, and this polynomial can be solved by factoring