Aptitude Discussion

Q. |
$F(x)$ is a fourth order polynomial with integer coefficients and with no common factor. The roots of $F(x)$ are –2, –1, 1, 2. If p is a prime number greater than 97, then the largest integer that divides $F(p)$ for all values of $p$ is |

✖ A. |
72 |

✖ B. |
120 |

✖ C. |
240 |

✔ D. |
360 |

**Solution:**

Option(**D**) is correct

Given that $F(x)=(x+2)(x+1)(x−1)(x−2)$

Putting $x = P$, we have $F(P)=(P+2)(P+1)(P−1)(P−2)$

Since $P$ is a prime number, $P$ is in the form $6K ± 1$, where $K$ is positive integer

$F(6K+1)=(6K+3)(6K+2)(6K)(6K−1)$

$=(36)(2K+1)(3K+1)(K)(6K−1) $-------- (1)

$F(6K−1)=(6K+1)(6K+2)(6K)(6K−3)$

$=36(6K+1)(3K+1)(K)(3K−1) $-------- (2)

Please note that the value of $K≥17$ and expression $F(6K+1)$ and $F(6K–1)$ always bear the factor 10.

Hence **360** is the correct choice.

Therefore option (D) is the correct choice

**Vikas**

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Hey Vikas,

Prime numbers are of the form $6K \pm 1$, but it's not vice-versa. i.e. all numbers resulting from $6K \pm 1$ are not prime numbers.

Also, it is important to know that NOT ALL prime numbers take the form of $6K \pm 1$.

In fact, there is no simple method to find ALL the prime numbers. Had, there been a way to find prime numbers, our lives would have been easier, given the contribution of prime numbers in our life.

**Sireesh**

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How can you say that $F(6k+1)$ and $F(6k-1)$ always have a factor of 10 for $k \geq 17$ ??

In $F(6K-1)$, the $(3K-1)$ term gives 50 which has a 10 factor. In $F(6K+1)$, the $(2K+1)(3K+1)$ evaluates to $35*52$ for $K=17$ which gives a ten factor because of the $5*2$.

Since $P$ is a prime number, $P$ is in the form $6K \pm1$, where $K$ is positive integer.

You mean $P=6K\pm 1$?

What if $K=16$?

Then, $P=97$ and $95$ and $95$ is not a prime number.

Correct me if I am wrong.