3 years ago

How do you factor x4-x3-7x2-x+6?

1 answer:

5 0

$x^4-x^3-7x^2-x+6=x^4-x^3-6x^2-x^2-x+6=\\ \\=x^2(x^2-x-6)-(x^2-x-6)=(x^2-1)(x^2-x-6)=\\ \\=(x-1)(x+1)(x^2-3x+2x-6)=\\ \\=(x-1)(x+1)[x(x-3)+2(x-3)]=(x-1)(x+1)(x-3)(x+2)$

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