Question

प्रश्न 1 निम्नलिखित व्यंजको में कौन-कौन से व्यंजक एक चर वाला बहुपद है? कारण के साथ बत 1 3x - 4x + 15 23-7 31+273 उत्तर1​

Answer 1

Answer:

fiNyh d{kkvksa esa] vki chth; O;atdksa vkSj muosQ tksM+] ?kVkuk] xq.kk vkSj Hkkx dk

vè;;u dj pqosQ gSaA ogk¡ vki ;g Hkh vè;;u dj pqosQ gSa fd fdl izdkj oqQN chth;

O;atdksa dk xq.ku[kaMu fd;k tkrk gSA vki fuEu chth; loZlfedkvksa vkSj mudk

xq.ku[kaMu esa mi;ksx dk iqu%Lej.k dj ldrs gSa%

(x + y)

2

= x

2

+ 2xy + y

2

(x – y)

2

= x

2

– 2xy + y

2

vkSj] x

2

– y

2

= (x + y) (x – y)

bl vè;k; esa] lcls igys ,d fo'ks"k izdkj osQ chth; O;atd dk] ftls cgqin

(polynomial) dgk tkrk gS] vkSj mlls lac¼ 'kCnkoyh (terminology) dk vè;;u djsaxsA

;gk¡ ge 'ks"kiQy izes; (Remainder Theorem), xq.ku[kaM izes; (Factor Theorem) vkSj

cgqinksa osQ xq.ku[kaMu esa buosQ mi;ksx dk Hkh vè;;u djsaxsA buosQ vfrfjDr] ge oqQN

vkSj chth; loZlfedkvksa dk vkSj oqQN fn, gq, O;atdksa dk xq.ku[kaMu djus rFkk eku

fudkyus osQ ckjs esa Hkh vè;;u djsaxsA

2.2 ,d pj okys cgqin

lcls igys ge ;kn djsaxs fd pj dks ,d izrhd ls izdV fd