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)=4P(x)=x2+bx+c is a factor of x4+6x2+25.x4+6x2+25=x4+10x2−4x2+25=(x4+10x2+25)−4x2=(x2+5)2−(2x)2⇒x4+6x2+25=(x2+2x+5)(x2−2x+5)One of them is a factor of 3x4+4x2+28x+5.3x4+4x2+28x+5=3x4−6x3+15x2+6x3−11x2+28x+5=3x2(x2−2x+5)+6x3−12x2+30x+x2−2x+5=3x2(x2−2x+5)+6x(x2−2x+5)+x2−2x+5⇒3x4+4x2+28x+5=(x2−2x+5)(3x2+6x+1)So, P(x)=x2−2x+5P(x)=0⇒Δ=22−4(5)=−16<0 has imaginary roots.P(1)=1−2+5=4

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