2. Polynomials
hard

અવયવ પાડો :  $x^{3}+13 x^{2}+32 x+20$

Option A
Option B
Option C
Option D

Solution

આપણે પ્રયત્નો દ્વારા જાણીએ કે $p(1)=0$ છે કે $p(-1)=0$ છે.

$p(x) =x^{3}+13 x^{2}+32 x+20 $

$\therefore p(1) =(1)^{3}+13(1)^{2}+32(1)+20$

$=(1)+13(1)+32(1)+20 $

$=1+13+32+20$

$\therefore p(1) =66 \neq 0 $

$p(x) =x^{3}+13 x^{2}+32 x+20 $

$\therefore p(-1) =(-1)^{3}+13(-1)^{2}+32(-1)+20$

$=(-1)+13(-1)+32(-1)+20$

$=(-1)+13-32+20$

$=-33+33$

$\therefore p(-1) =0 $

અહીં, $p(-1) =0 $ શૂન્ય છે તેથી અવયવ પ્રમેયને આધારે $[x-(-1)]$ એટલે $x + 1$ એ $p(x)$ નો અવયવ છે.

$\therefore x^{3}+13 x^{2}+32 x+20 =(x+1)\left(x^{2}+12 x+20\right) $

$=(x+1)\left[x^{2}+2 x+10 x+20\right] $

$=(x+1)[x(x+2)+10(x+2)] $

$=(x+1)(x+2)(x+10)$

Standard 9
Mathematics

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