Question

Q) Division of a polynomial by monomial :

8(x³y²z²+x²y³z²+x²y²z³)÷2x²y²z²
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Answer 1

Answer:

2(x+y+z)

Step-by-step explanation:

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Answer 2

Answer :

• 4 ( x + y + z )

Question :

• 8(x³y²z²+x²y³z²+x²y²z³) ÷ 2x²y²z²

To do :

• Division of a polynomial by monomial.

Using :

1. Distributive Law.
2. Normal Divide.

Solution :

Using Formula no. 1 :

$\rm\red{8(x {}^{3} y {}^{2} z {}^{2} +x {}^{2} y {}^{3} z {}^{2} +x {}^{2} y {}^{2} z {}^{3} )\div2x {}^{2} y {}^{2} z {}^{2} } \\ \\ = \rm8(x {}^{3} y {}^{2} z {}^{2} +x {}^{2} y {}^{3} z {}^{2} +x {}^{2} y {}^{2} z {}^{3})\div \frac{1}{2x{}^{2}y{}^{2}z{}^{2}} \\ \\ \rm = \frac {8x{}^{3}y{}^{2}z{}^{2}}{2x{}^{2}y{}^{2}z{}^{2}} + \frac{8x{}^{2}y{}^{3}z{}^{2}}{2x{}^{2}y{}^{2}z{}^{2}} + \frac{8x{}^{2}y{}^{2}z{}^{3}}{2x{}^{2}y{}^{2}z{}^{2}} \\ \\ \rm = 4x + 4y + 4z \\ \\ \large \rm \green{ = 4(x + y + z)}$

Using Formula no. 2 :

$\rm\pink{8(x {}^{3} y {}^{2} z {}^{2} +x {}^{2} y {}^{3} z {}^{2} +x {}^{2} y {}^{2} z {}^{3} )\div2x {}^{2} y {}^{2} z {}^{2}} \\ \\ \gray{ \bigg( 8(x{}^{3}y{}^{2}z{}^{2}+x{}^{2}y{}^{3}z{}^{2}+x{}^{2}y{}^{2}z{}^{3}) =8x{}^{2}y{}^{2}z{}^{2}(x + y + z) \bigg)} \\ \\ \rm = \frac {8x{}^{2}y{}^{2}z{}^{2}(x + y + z)}{2x{}^{2}y{}^{2}z{}^{2}} \\ \\ \large \rm \blue{ = 4(x + y + z)}$

ㅤ∴ The Answer Is :

ㅤㅤㅤㅤ$\huge \bigg[ \small \rm\purple{4 ( x + y + z )}\bigg]$