Roots

3 solutions:

Solution 1:

In the following expansion, we'll replace r²+2r+6 with 0.

(r+2)(r+3)(r+4)(r+5) = (r²+5r+6)(r²+9r+20) = (r²+2r+6+3r)(r²+2r+6+7r+14) = 3r(7r+14) = 21(r²+2r) = 21(r²+2r+6)-21*6 = -126

Solution 2:

if r is the root of x^2 + 2x + 6 =0 then, r^2 + 2r + 6 is also = to 0.

Evaluate first:

(r+2)(r+3)(r+4)(r+5) = (r^2 + 2r + 6 +3r)(r^2 + 2r + 7r + 6 + 14)

= (3r)(7r + 14)

= 21(r^2 + 2r)

= 21(-6)

= -126

Solution 3:

If r is a root of x^2 + 2x + 6 = 0, (x-r)(x-6/r) = 0 ---> x^2 - x(r+6/r) + 6 = 0 = x^2 + 2x + 6 ---> -x(r+6/r) = 2x ----> r+6/r = -2 -----> r^2 + 2r = -6

then, (r+2)(r+3)(r+4)(r+5) = (r^2)(r+ 6/r + 5)(r+6/r + 14/r + 9)

= (r^2)(5-2)(9-2 + 14/r)

= (3r)(7r+14)

= 21r^2 + 42r

= 21(r^2 + 2r)

= 21(-6)

= -126

$x^{2} + 2x + 1 + 5 =0$

$(x + 1)^{2} = -5$

$x = -1 \pm i\sqrt{5}$

( since any root is asked)

thus

$(- 1 + i\sqrt{5} + 2)( - 1 + i\sqrt{5} + 3)( - 1 + i\sqrt{5} + 4)( - 1 + i\sqrt{5} + 5)$

$= (1 + i\sqrt{5})(2 + i\sqrt{5})(3 + i\sqrt{5})(4 + i\sqrt{5})$

$= (1 + i\sqrt{5})(4 + i\sqrt{5})(2 + i\sqrt{5})(3 + i\sqrt{5})$

$= ( -1 + 5i\sqrt{5})(1 + 5i\sqrt{5})$

$= -126$

$r$ is a root of the equation $x^2 + 2x + 6 = 0$ ,

$\Rightarrow r^2 + 2r + 6 = 0$

$\Rightarrow r^2 + 2r= - 6$ $\rightarrow$ we can replace $r^2+2r$ with $-6$

Therefore,

$(r+2)(r+3)(r+4)(r+5)$

$=(r+2)(r+5)(r+3)(r+4)$

$=(r^2+7r+10)(r^2+7r+12)$

$=[(r^2+2r)+5r+10] [_bold_(r^2+2r)+5r+12]$

$=({-6}+5r+10)({-6}+5r+12]$

$=(5r+4)(5r+6$

$=25r^2+50r+24$

$=25(r^2+2r)+24$

$=25(-6)+24$

$=-150+24$

$=\boxed{126}$