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Question

Let P(x)=x2+bx+c, where b and c are integer. If P(x) is a factor of both x4+6x2+25 and 3x4+4x2+28x+5, then

A
P(x)=0 has imaginary roots
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B
P(x)=0 has roots of opposite sign
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C
P(1)=4
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D
P(1)=6
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Solution

The correct options are
A P(x)=0 has imaginary roots
B P(1)=4

P(x)=x2+bx+c is a factor of x4+6x2+25.

x4+6x2+25=x4+10x24x2+25=(x4+10x2+25)4x2=(x2+5)2(2x)2

x4+6x2+25=(x2+2x+5)(x22x+5)

One of them is a factor of 3x4+4x2+28x+5.

3x4+4x2+28x+5=3x46x3+15x2+6x311x2+28x+5=3x2(x22x+5)+6x312x2+30x+x22x+5=3x2(x22x+5)+6x(x22x+5)+x22x+5

3x4+4x2+28x+5=(x22x+5)(3x2+6x+1)

So, P(x)=x22x+5

P(x)=0Δ=224(5)=16<0 has imaginary roots.

P(1)=12+5=4


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