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Question

If the zeroes of p(x) are double the zeroes of the polynomial q(x) = x24x+3, then p(x) is

A
x28x+12
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B
2x28x+4
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C
x24x+3
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D
4x24x+8
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Solution

The correct option is A x28x+12
Splitting the middle term of q(x), we have,
x24x+3
x2x3x+3
x(x1)3(x1)
(x3)(x1)
The zeroes of q(x) are 1 and 3.
We know, zeroes of p(x) are double that of q(x).
The zeroes of p(x) are 2 and 6.
p(x)=(x2)(x6)
p(x)=x28x+12​​​​

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