Question

phân tích đa thức thành nhân tử (x^2+x+1)(x^2+x+2)-12

Answer 1

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

Answer 2

(x^2+x+1)(x^2+x+2)-12

Đặt x^2+x+1= a ta có

=a^2+a-12

=a^2-3a+4a-12

=(a^2-3a)+(4a-12)

=a(a-3)+4(a-3)

=(a-3)(a+4)

thay x^2+x+1=a ta được

(x^2+x-2)(x^2+x+5)

Answer 3

                                         Bài làm :

Ta có :

(x²+x+1)(x²+x+2)-12

=(x²+x+1)[(x²+x+1)+1)-12

=(x²+x+1)²+(x²+x+1)-12

=(x²+x+1)²-3.(x²+x+1)+4.(x²+x+1)-12

=(x²+x+1)(x²+x+1-3)+4.(x²+x+1-3)

=(x²+x+1)(x²+x-2)+4.(x²+x-2)

=(x²+x-2)(x²+x+1+4)

=(x²-x+2x-2)(x²+x+5)

=[x.(x-1)+2.(x-1)](x²+x+5)

=(x-1)(x+2)(x²+x+5)

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