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2014-12-18T12:28:57+05:30

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(x+1)^6-(x-1)^6\\\\=[(x+1)^3]^2-[(x-1)^3]^2\\\\=[(x+1)^3+(x-1)^3]*[(x+1)^3-(x-1)^3]\\\\=[2x^3+6x]*[6x^2+2]\\\\=2x(x+3)2(3x^2+1)\\


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P(x)=(x^2-3x)(x^2-3x+7)+10\\\\=(x^2-3x)^2+7x^2-21x+10\\\\=x^4-6x^3+16x^2-21x+10

On examining coefficients, rational roots are possibly factors of +10 or -10, ie.,+1,+2,+5,+10,-1,-2,-5-10.

Checking with x =1, P(1)  = 0,  Hence (x-1) is a factor of P(x),
Checking with x = 2, P(2)=0, Hence (x-2) is a factor too.

P(x) = (x-1)(x-2)(x^2+ax+5) (x^2-3x+2)(x^2+ax+5)

Coefficient of x^3 will be -3+a = -6  ,  hence a = -3

Factors are : P(x) = (x-1)(x-2)(x^2-3x+5)
the quadratic expression has imaginary roots, as 3^2-4*5 <0.  Hence there are no real factors.



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