What are all the possible rational zeros for #f(x)=2x^3+x^2+8x-21# and how do you find all zeros?
1 Answer
Explanation:
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/2, +-1, +-3/2, +-3, +-7/2, +-7, +-21/2, +-21#
In addition note that the signs of the coefficients are in the pattern
So let's try the positive rational possibilities first:
#f(1/2) = 2(1/8)+(1/4)+8(1/2)-21 = -33/2#
#f(1) = 2+1+8-21 = -10#
#f(3/2) = 2(27/8)+(9/4)+8(3/2)-21 = (27+9+48-84)/4 = 0#
So
#2x^3+x^2+8x-21 = (2x-3)(x^2+2x+7)#
We can factor the remaining quadratic by completing the square and using the difference of squares identity:
#a^2-b^2 = (a-b)(a+b)#
with
#x^2+2x+7 = x^2+2x+1+6#
#color(white)(x^2+2x+7) = (x+1)^2-(sqrt(6)i)^2#
#color(white)(x^2+2x+7) = ((x+1)-sqrt(6)i)((x+1)+sqrt(6)i)#
#color(white)(x^2+2x+7) = (x+1-sqrt(6)i)(x+1+sqrt(6)i)#
Hence zeros:
#x = -1+-sqrt(6)i#