Perform the following division and
verify the result (3p²+6p+18)devide by p-6
Answers
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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Answer:
p-6/3p^2+6p+18
p-6✓3p^2+6p+18 =3p+24
3p^2-18p
24p+18
24p-144
-162