Question

Divide 3y⁵ + 6y⁴ + 6y³ +7y² + 8y + 9 by 3y³ + 1 also verify the results ​

Answer 1

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Dividend = 3y⁵ + 6y⁴ + 6y³ +7y² + 8y + 9

Divisor = 3y³ + 1

Quotient = y² + 2y + 2

Remainder = 6y² + 6y + 7

We know that:-

∴ Dividend = Divisor × Quotient + Remainder

$\red↦$3y⁵ + 6y⁴ + 6y³ +7y² + 8y + 9 = (3y³ + 1)×(y² + 2y + 2) + (6y² + 6y + 7)

$\red↦$3y⁵ + 6y⁴ + 6y³ +7y² + 8y + 9 = 3y⁵ + 6y⁴ + 6y³ + y² + 2y + 2 + (6y² + 6y + 7)

$\red↦$3y⁵ + 6y⁴ + 6y³ +7y² + 8y + 9 = 3y⁵ + 6y⁴ + 6y³ + y² + 6y² + 2y + 6y + 2 + 7

$\red↦$3y⁵ + 6y⁴ + 6y³ +7y² + 8y + 9 = 3y⁵ + 6y⁴ + 6y³ +7y² + 8y + 9

$\red{\therefore}$ As LHS = RHS (Hence Verified)

Answer 2

this answer is in DAV public School secondary mathematics book of class 8 page number 135

Step-by-step explanation:

quotient to will be y² + 2 y + 2

remainder will bep 6 y² + 6 y + 7