Math, asked by prakasharora6453, 9 months ago

Perform the following division and
verify the result (3p²+6p+18)devide by p-6​

Answers

Answered by hodeee4
5

Answer:

  • 162

Step-by-step explanation:

According to question,

According to question,p−6

According to question,p−6(3p

According to question,p−6(3p 2

According to question,p−6(3p 2 +6p+18)

According to question,p−6(3p 2 +6p+18)

According to question,p−6(3p 2 +6p+18)

According to question,p−6(3p 2 +6p+18) ⟹

According to question,p−6(3p 2 +6p+18) ⟹ p−6

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54]

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54]

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54]

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54] ⟹3(p+8)+

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54] ⟹3(p+8)+ (p−6)

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54] ⟹3(p+8)+ (p−6)162

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54] ⟹3(p+8)+ (p−6)162

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54] ⟹3(p+8)+ (p−6)162

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54] ⟹3(p+8)+ (p−6)162 Here, Numerator is not completely divisible by denominator.

According to question,p−6(3p 2 +6p+18) ⟹ p−63[(p−6)(p+8)+54] ⟹3(p+8)+ (p−6)162 Here, Numerator is not completely divisible by denominator.Hence, remainder is 162

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Answered by rehabgyanpeethika
7

Answer:

p-6/3p^2+6p+18

p-6✓3p^2+6p+18 =3p+24

3p^2-18p

24p+18

24p-144

-162

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