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Question

The H.C.F. of x33x2+x3 and x4+6x2+5 is .

A
x4+6x2+5
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B
x+3
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C
x3
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D
x2+1
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Solution

The correct option is D x2+1
Here we do not find any common terms, so we simply divide.
x33x2+x3)x4 +6x2+5(x+3 x43x3+x23x + +––––––––––––––– 3x3+5x2+3x+5 3x39x2+3x9 + +–––––––––––––– 14x2 +14 =14(x2+1)
Here the highest power of divisor is 3 and that of remainder is 2.
So now remainder becomes divisor and divisor becomes the dividend.
x2+1)x33x2+x3(x3 x3 +x ––––––––––– 3x2 3 3x2 3–––––––––––– ×
Now the remainder in zero. So the required H.C.F is the divisor i.e. x2+1

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