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:

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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