Question

Q45. Which one of the following is TRUE?
O (A) x-2 is reducible in Q[x]
0 (B) x2+1 is irreducible in C[x]
0 (C) x²+25 is reducible in C[x]
O (D) x?1 is irreducible in RAX​

Answer 1

ʜɪɪɪ

ɢᴏɪɴɢ ᴄᴀsʜʟᴇss ɴᴏᴛ ᴏɴʟʏ ᴇᴀsᴇs ᴏɴᴇ's ʟɪғᴇ ʙᴜᴛ ᴀʟsᴏ ʜᴇʟᴘs ᴀᴜᴛʜᴇɴᴛɪᴄᴀᴛᴇ ᴀɴᴅ ғᴏʀᴍᴀʟɪᴢᴇ ᴛʜᴇ ᴛʀᴀɴsᴀᴄᴛɪᴏɴs ᴛʜᴀᴛ ᴀʀᴇ ᴅᴏɴᴇ. ᴛʜɪs ʜᴇʟᴘs ᴛᴏ ᴄᴜʀʙ ᴄᴏʀʀᴜᴘᴛɪᴏɴ ᴀɴᴅ ᴛʜᴇ ғʟᴏᴡ ᴏғ ʙʟᴀᴄᴋ ᴍᴏɴᴇʏ ᴡʜɪᴄʜ ʀᴇsᴜʟᴛs ɪɴ ᴀɴ ɪɴᴄʀᴇᴀsᴇ ᴏғ ᴇᴄᴏɴᴏᴍɪᴄ ɢʀᴏᴡᴛʜ. ᴛʜᴇ ᴇxᴘᴇɴᴅɪᴛᴜʀᴇ ɪɴᴄᴜʀʀᴇᴅ ɪɴ ᴘʀɪɴᴛɪɴɢ ᴀɴᴅ ᴛʀᴀɴsᴘᴏʀᴛᴀᴛɪᴏɴ ᴏғ ᴄᴜʀʀᴇɴᴄʏ ɴᴏᴛᴇs ɪs ʀᴇᴅᴜᴄᴇᴅ.

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

Which one of the following is TRUE?

(A) x-2 is reducible in Q[x]

(B) x² + 1 is irreducible in C[x]

(C) x² + 25 is reducible in C[x]

(D) x² - 1 is irreducible in R[x]

EVALUATION

Here x - 2 can be rewritten as the product of polynomials whose degree is less than the given polynomial

So x-2 is irreducible in Q[x]

Now

$\sf{ {x}^{2} + 1 }$

$= \sf{(x + i)(x - i)}$

So x² + 1 is reducible in C[x]

Again

$\sf{ {x}^{2} + 25}$

$\sf{ = (x + 5i)(x - 5i)}$

Hence x² + 25 is reducible in C[x]

Again

x² - 1

= (x+1)(x-1)

Hence x² - 1 is reducible in R[x]

FINAL ANSWER

The correct option is

(C) x² + 25 is reducible in C[x]

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