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Factorise `x^(3)-7x+6`. |
Answer» Here, constant term=6 Factors of `6=+-1, +-2,+-3,+-6` Let `p(x)=x^(3)-7x+6` Put x=1 Remainder `=p(1)=1^(3)-7(1)+6=1-7+6=0` `therefore (x-1)`, is factor of p(x). Now, `p(x)=x^(3)-7x+6=x^(2)(x-1)+x^(2)-7x+6` `=x^(2)(x-1)+x(x-1)-6x+6` `=x^(2)(x-1)+x(x-1)-6(x-1)` `=(x-1)(x^(2)+x-6)` `=(x-1)[x^(2)+3x-2x-6]` `=(x-1)[x(+3)-2(x+3)]` =(x-1)(x+3)(x-2) |
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