What are the possible number of positive real, negative real, and complex zeros of #f(x) = 4x^3 + x^2 + 10x – 14#?

Question

5896 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-26T14:25:58

This cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

#f(x) = 4x^3+x^2+10x-14#

The signs of the coefficients follow the pattern: #+ + + -#

With one change of sign, we can tell that this cubic polynomial has one positive zero.

Consider #f(-x)#...

#f(-x) = -4x^3+x^2-10x-14#

The signs of the coefficients follow the pattern: #- + - -#

With two changes of sign, we can tell that this cubic has #0# or #2# negative zeros.

Let us examine the discriminant:

#4x^3+x^2+10x-14# is in the form #ax^3+bx^2+cx+d# with #a=4#, #b=1#, #c=10# and #d=-14#.

This has discriminant #Delta# given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

#=100-16000+56-84672-10080#

#=-110596#

Since #Delta < 0# we can deduce that #f(x)# has one Real zero and a pair of Complex conjugate zeros.

graph{4x^3+x^2+10x-14 [-10, 10, -200, 200]}