[tex](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)\\[/tex]
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[tex]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[/tex]
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.
