Divide the polynomial by monomial:
$y^{3} - 64; y - 4$

(y - 4) (y^{2} + 4 y + 16) + 0

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(y + 2) (y^{2} + y + 16) + 0

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(y + 74) (y^{2} - 4 y + 16) + 0

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(y - 3) (y^{2} + 3 y + 2) + 0

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Solution

The correct option is A $(y - 4) (y^{2} + 4 y + 16) + 0$
$y^{3} - 64 \div y - 4$
$y^{2} + 4 y + 16$
$y - 4$ $y^{3} - 64$
$y^{3} - 4 y^{2}$
- +
$+ 4 y^{2} - 64$
$+ 4 y^{2} - 16 y$
- +
$16 y - 64$
$16 y - 64$
- +
$0$
Therefore,l
$y^{3} - 64 = (y - 4) (y^{2} + 4 y + 16) + 0$

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Similar questions

(3 y^{8} - 4 y^{6} + 5 y^{4}) \div y^{4}

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x -

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y^{2} = 4 x

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(3 y^{8} - 4 y^{6} + 5 y^{4}) \div y^{4}

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