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The factors of the expression $27p³ – 1216 – 92 p²+ 14 p$ is:

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(p –

16

) (p –

16

) (p –

16

)

(2p –

16

) (2p –

16

) (2p –

16

)

(3p –

16

) (3p –

16

) (3p –

16

)

(4p –

16

) (4p –

16

) (4p –

16

)

answer is C.

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Detailed Solution

Given:

27p³ – 1216 – 92 p²+ 14 p

Now, let us rewrite the given expression as :

27 p 3 − 1216 − 92 p 2 + 14 p = (3 p) 3 − (16) 3 − 3 (3 p) 2 (16) + 3 (3 p) (16) 2

Let us use the identity here:
(a – b)³ = a³ – b³ – 3a²b + 3ab²

= (3 p − 16) 3

= (3 p − 16) (3 p − 16) (3 p − 16)

Hence,

27p³ – 1216 – 92 p²+ 14 p

= (3p –

16

) (3p –

16

) (3p –

16

)
Therefore, the correct option is 3.

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The factors of the expression 27p³ – 1 216 – 9 2 p²+ 1 4 p is:

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The factors of the expression $27p³ – 1216 – 92 p²+ 14 p$ is: