How do you solve #(x+5)(x-2)(x-1)(x+1)<0#?

1 Answer
Aug 12, 2016

Answer:

#x in (-5, -1) uu (1, 2)#

Explanation:

Quick method

The left hand side is an upright quartic (positive coefficient of #x^4#) which intersects the #x# axis at the four zeros: #-5, -1, 1, 2#.

So it will be negative precisely in the intervals #(-5, -1)# and #(1, 2)#.

Here's a graph of #y = 1/20 (x+5)(x-2)(x-1)(x+1)# ...

graph{(x+5)(x+1)(x-1)(x-2)/20 [-10, 10, -5.5, 4.5]}

#color(white)()#
A little slower...

Let #f(x) = (x+5)(x-2)(x-1)(x+1)#

#f(x)# is a continuous function which is zero when #x in { -5, -1, 1, 2 }#, so does not satisfy the inequality #f(x) < 0# at those points.

These points divide the Real line into #5# intervals in which the sign of #f(x)# cannot change:

#(-oo, -5), (-5, -1), (-1, 1), (1, 2), (2, oo)#

Since the zeros are each of multiplicity #1# - which is an odd number - the sign does change at each of these points.

When #x < -5#, the sign of each of the linear factors is negative. When they are all multiplied, the result is positive.

That is, when #x < -5#:

#{ ((x+5) < 0), ((x-2) < 0), ((x-1) < 0), ((x+1) < 0) :}#

Hence #(x+5)(x-2)(x-1)(x+1) > 0#

So #f(x)# has the following signs in each of the #5# intervals:

#+, -, +, -, +#

respectively.

So #f(x) < 0# in #(-5, -1) uu (1, 2)#