How do you solve #x^3+7x^2+10x>=0#?

1 Answer
Oct 1, 2016

Answer:

Solution is #-5 >= x>=-2# or #x >= 0#

Explanation:

Let us first factorize #x^3+7x^2+10x#.

#x^3+7x^2+10x=x(x^2+7x+10)=x(x^2+2x+5x+10)#

= #x(x(x+2)+5(x+2)=x(x+2)(x+5)#

Hence we have to solve the inequality #x^3+7x^2+10x>0# or #(x+5)(x+2)x>=0#

From this we know that for the product #(x+5)(x+2)x>=0#, signs of binomials #(x+5)#, #(x+2)# and #x# will change around the values #-5#. #-2# and #0# respectively. In sign chart we divide the real number line around these values, i.e. below #-5#, between #-5# and #-2#, between #-2# and #0# and above #0# and see how the sign of #(x+5)(x+2)x# changes.

Sign Chart

#color(white)(XXXXXXXXXXX)-5color(white)(XXXXX)-2color(white)(XXXXX)0#

#(x+5)color(white)(XXXX)-ive color(white)(XXXX)+ive color(white)(XX)+ive color(white)(XXX)+ive#

#(x+2)color(white)(XXXX)-ive color(white)(XXXX)-ive color(white)(XX)-ive color(white)(XXX)+ive#

#xcolor(white)(XXXXXXX)-ive color(white)(XXXX)-ive color(white)(XX)+ive color(white)(XXX)+ive#

#(x+5)(x+2)x#
#color(white)(XXXXXXXX)-ive color(white)(XXXX)+ive color(white)(XX)-ive color(white)(XXX)+ive#

It is observed that #(x+5)(x+2)x>= 0# when either #-5 >= x>=-2# or #x >= 0#, which is the solution for the inequality.