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Question

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

A
a+b+x
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B
abx
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C
ab+x
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D
a+bx
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Solution

The correct option is D a+bx
Using C1C1+C2+C3, we get
Δ=∣ ∣a+bxaba+bxxaa+bxbx∣ ∣=(a+bx)∣ ∣1ab1xa1bx∣ ∣=(a+bx)(x2+(a+b)x+a2+b2ab)

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