y³ $-$ 6y² + 6y + 1 ; y $-$ 1

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Solution

Here, quotient = y² $-$ 5y + 1
and remainder = 2

$∴ y^{3} - 6 y^{2} + 6 y + 1 = (y - 1) (y^{2} - 5 y + 1) + 2$

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Similar questions

Q. Divide the polynomial by monomial:

y^{3} - 6 y^{2} + 6 y + 1; y - 1

Q. Simplify:

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, then the answer is

y^{3} - 6 y^{2} + 5 y

If true then enter

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and if false then enter

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y3 - 6y2 + 6y + 1 ; y - 1

Here, quotient = y2 - 5y + 1 and remainder = 2 ∴y3-6y2+6y+1=y-1y2-5y+1+2

y³ $-$ 6y² + 6y + 1 ; y $-$ 1

Here, quotient = y² $-$ 5y + 1
and remainder = 2

$∴ y^{3} - 6 y^{2} + 6 y + 1 = (y - 1) (y^{2} - 5 y + 1) + 2$