Let $r$ be a root of the equation $x^{2} + 2 x + 6 = 0$ . The value of $(r + 2) (r + 3) (r + 4) (r + 5)$ is equal to.

51

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- 51

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- 126

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126

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Solution

The correct option is B $- 126$
$r$ is a root of the equation $x^{2} + 2 x + 6 = 0$

$∴ r^{2} + 2 r + 6 = 0$

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

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

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

$= (r^{2} + 2 r + 6 + 5 r + 4) (r^{2} + 2 r + 6 + 5 r + 6)$

$= (5 r + 4) (5 r + 6)$ ....................... $(∵ r^{2} + 2 r + 6 = 0)$

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

$= 25 (- 2 r - 6) + 50 r + 24$

$= - 50 r - 150 + 50 r + 24$

$= - 126$

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Let r be a root of the equation x2+2x+6=0. The value of (r+2)(r+3)(r+4)(r+5) is equal to.

The correct option is B −126r is a root of the equation x2+2x+6=0∴r2+2r+6=0(r+2)(r+3)(r+4)(r+5)=(r+2)(r+5)(r+3)(r+4)=(r2+7r+10)(r2+7r+12)=(r2+2r+6+5r+4)(r2+2r+6+5r+6)=(5r+4)(5r+6).......................(∵r2+2r+6=0)=25r2+50r+24=25(−2r−6)+50r+24=−50r−150+50r+24=−126

Let $r$ be a root of the equation $x^{2} + 2 x + 6 = 0$ . The value of $(r + 2) (r + 3) (r + 4) (r + 5)$ is equal to.