Question

$6y {3}^{} - 7y {}^{2} - 8y + 5$
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Answer 1

Answer:

56

Step-by-step explanation:

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Answer 2

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$</p><p>(y \: + 1) \: (2y \: - 1) \: (3y \: - \: 5)$

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$p \: (y) = \: 6y3 \: - 7y2 \: - 8y + 5$$p \: ( - ) \:</p><p> = 6 \: - 7 + 8 + 5$

$= \: 13 \: + 13 \: = \: 0$$p \: ( \: - 1) \: 0 = \ </p><p>\:: : \implies \: y \: + \: 1 \: is \: a \: factor \: of \: p \: (y)$$\ \: : </p><p>\implies \: \: 6y {}^{3} \: - \: 7y {}^{2} \: - \: 8y \: + 5$$= \: (y \: + \: 1 \: ) \: (6y {}^{2} \: - 13y \: + \: 5)$$6y \: {}^{2} \: - \: 13y \: + \: 5 = \: 6y {}^{2} \: - \: 3y \: - \: 10y \: + 5 \: = 3y \: (2y - \: 1 \: ) \: - 5(2y - 1)$$= \: (3y - \: 5) \: (2y \: - \: 1)$$\ \: : \implies \: \: 6y {}^{3} \: - \: 7y {}^{2} \: - \: 8y \: + \: 5$$\ \: :</p><p> \implies \: \: (y + 1) \: (3y - 5) \: (2y - 1)$