Question

Phân tích đa thức thành nhân tử bằng phương pháp đặt biến phụ dạng (x+a)(x+b)(x+c)(x+d) + ex² ( với ab=cd)
4(x+5)(x+6)(x+10)(x+12) - 3x²

Answer 1

\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)

\(=2\left(x+5\right)\left(x+12\right).2\left(x+6\right)\left(x+10\right)-3x^2\)

\(=\left(2x^2+34x+120\right).\left(2x^2+32x+120\right)-3x^2\)

Đặt: \(a=2x^2+33x+120\) , ta có:

\(\left(a+x\right)\left(a-x\right)-3x^2\)

\(=a^2-x^2-3x^2\)

\(=a^2-4x^2\)

\(=\left(a-2x\right)\left(a+2x\right)\)

Thay \(a=2x^2+33x+120\) ta có:

\(\left(2x^2+33x+120-2x\right)\left(2x^2+33x+120+2x\right)\)

\(=\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

\(=\left(2x^2+16x+15x+120\right)\left(2x^2+35x+120\right)\)

\(=\left[2x\left(x+8\right)+15\left(x+18\right)\right]\left(2x^2+35x+120\right)\)

\(=\left(x+8\right)\left(2x+15\right)\left(2x^2+35x+120\right)\)

Answer 2

4(x+5)(x+6)(x+10)(x+12) - 3x²
= 2[(x+5)(x+12)] . 2[(x+6)(x+10)] - 3x²
= 2(x²+60+17x) . 2(x²+60+16x) - 3x²
= (2x²+120+34x)(2x²+120+32x) - 3x²
= (2x²+120+33x + x)(2x²+120+33x - x) - 3x²
= (2x²+120+33x) - x² - 3x²
= (2x²+120+33x) - 4x²
= (2x²+120+33x - 2x)(2x²+120+33x + 2x)
= (2x²+120+31x)(2x²+120+35x)
= (2x²+15x+16x+120)(2x²+35x+120)
= [2x(x+8) + 15(x+8)](2x²+35x+120)
= (x+8)(2x+15)(2x²+35x+120)