perform the following division and give the remainder.
1. (3p^2+6p+18)÷(p-6)
2. (a^2+2ab+b^2)÷(a+b)
Answers
Answered by
2
=(3p2+6p+18)÷(p−6)
= \frac{3p {}^{2} + 6p + 18 }{p - 6}=p−63p2+6p+18
= (a {}^{2} + 2ab + b {}^{2} ) \div (a + b)=(a2+2ab+b2)÷(a+b)
= \frac{a {}^{2} + 2ab + b {}^{2} }{a + b}=a+ba2+2ab+b2
= \frac{(a + b {}^{2}) }{a + b}=a+b(a+b2)
= a + b=a+b
this may help you
= \frac{3p {}^{2} + 6p + 18 }{p - 6}=p−63p2+6p+18
= (a {}^{2} + 2ab + b {}^{2} ) \div (a + b)=(a2+2ab+b2)÷(a+b)
= \frac{a {}^{2} + 2ab + b {}^{2} }{a + b}=a+ba2+2ab+b2
= \frac{(a + b {}^{2}) }{a + b}=a+b(a+b2)
= a + b=a+b
this may help you
Similar questions