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Question

State True or False: x32x2x+2 = (x1)(x2)(x+1)

A
True
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B
False
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Solution

The correct option is A True
Let p(x)=x32x2x+2
By trial, we find that
p(1)=(1)32(1)2(1)+2
=121+2=0
By factor Theorem (x-1) is a factor of p(x)
Now, x32x2x+2
=x3x2x2+x2x+2
=x2(x1)x(x1)2(x1)
=(x1)(x2x2)
=(x1)(x22x+x2)
=(x1){x(x2)+1(x2)}
=(x1)(x2)(x+1)

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