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Question

If Δ=∣ ∣xabbxaabx∣ ∣ then a factor of Δ is

A
a+b+x
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B
x2(ab)x+a2+b2+ab
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C
x2+(a+b)x+a2+b2ab
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D
a+bx
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Solution

The correct options are
B x2+(a+b)x+a2+b2ab
C a+bx
Applying C1C1+C2+C3
Δ=∣ ∣a+bxaba+bxxaa+bxbx∣ ∣=(a+bx)∣ ∣1ab1xa1bx∣ ∣
Again applying R2R2R1,R3R3R1
Δ=(a+bx)∣ ∣1ab0xaab0baxb∣ ∣=(a+bx)((x+a)(x+b)+(ab)2)=(a+bx)(x2+(a+b)x+a2+b2ab)

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