Question

Let p(x)=x2+bx+c where b and c are integers. If p(x) is a factor of both x4+6x2+25 and 3x4+4x2+28x+5, what is p(1) ?

A
0
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B
1
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C
2
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D
4
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Solution

The correct option is B 4If p(x) is the common factor of both x4+6x2+25 and 3x4+4x2+28x+5 then it means that p(x) is the HCF of both the expressions. So, we divide 3x4+4x2+28x+5 by x4+6x2+25 as shown in the above image:Therefore, the remainder is −14x2+28x−70 that is R(x)=−14(x2−2x+5).Thus, the HCF is x2−2x+5 which means that p(x)=x2−2x+5Now, substitute x=1 in p(x)=x2−2x+5 as shown below:p(1)=12−(2×1)+5=1−2+5=6−2=4Hence, p(1)=4.

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