Algebra

Quantitative Aptitude (RRB NTPC)

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x³ + 6x² + 11x + 6 is divisible by

Only (x + 1)

Only (x + 2)

Only (x + 3)

All of these

Explanation:

Let f(x) = x³ + 6x² + 11x + 6
f(x) = 0. So, put x = – 1, –2 and –3
f(–1) = (–1)³ + 6(–1)²+ 11(–1) + 6 = –1 + 6 – 11 + 6
= –12 + 12 = 0.
f(–2) = (–2)³ + 6(–2)²+ 11(–2) + 6 = –8 + 24 – 22 + 6
= –30 + 30 = 0.
f(–3) = (–3)³ + 6(–3)²+ 11(–3) + 6 = –27 + 54 – 33 + 6
= –60 + 60 = 0.
Hence, (x + 1), (x + 2) and (x + 3) are the factors of f(x).

By: Amit Kumar

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