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Perform the following division andverify the result (3p²+6p+18)devide by p-6 |
Answer»
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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