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

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

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Answer 1 · 2018-02-28T11:39:16

Please see the handscript below ; $- 126$ $-126$

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Answer 2 · 2018-02-28T11:39:29

$- 126$ $-126$

Explanation:

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

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

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