Let #r# be the root of the equation #x^2 + 2x + 6#.What is the value of #(r+2)(r+3)(r+4)(r+5)#?

2 Answers
Feb 28, 2018

Please see the handscript below ; #-126#

Explanation:

answer
Hope this helps

Feb 28, 2018

#-126#

Explanation:

The equationmust be
#x^2+2x+6=color(red)0#
If r is one of the root of this equation,then
#r^2+2r+6=0rArrcolor(red)(r^2+2r)=-6,and color(red)(r^2)=-2r-6#
Now, #(r+2)(r+3)(r+4)(r+5)=(color(red)(r^2)+5r+6)(color(red)(r^2)+9r+20)=(-2r-6+5r+6)(-2r-6+9r+20)##=(3r)(7r+14)=21r(r+2)=21(color(red)(r^2+2r))=21(-6)#
#=-126#