Đáp án:
Giải thích các bước giải:
f/ $2x^3+3x^2-5x-6$
$=2x^3+2x^2+x^2+x-6x-6$
$=2x^2(x+1)+x(x+1)-6(x+1)$
$=(x+1)(2x^2+x-6)$
$=(x+1)(2x^2+4x-3x-6)$
$=(x+1)[2x(x+2)-3(x+2)]$
$=(x+1)(x+2)(2x-3)$
g/ $3x^3-10x^2+9$ (đề được sửa dưới phần bình luận !)
$=3x^3-9x^2-x^2+9$
$=3x^2(x-3)-(x^2-9)$
$=3x^2(x-3)-(x-3)(x+3)$
$=(x-3)(3x^2-x-3)$
$=\dfrac{1}{12}.(x-3)(36x^2-12x-36)$
$=\dfrac{1}{12}.(x-3)[(36x^2-12x+1)-37]$
$=\dfrac{1}{12}.(x-3)[(6x-1)^2-37]$
$=\dfrac{1}{12}.(x-3)(6x-1-\sqrt{37})(6x-1+\sqrt{37})$
k/ $4x^4-25x^2+36$
$=(2x^2)^2-2.6.2x^2+36-x^2$
$=(2x^2-6)^2-x^2$
$=(2x^2-6-x)(2x^2-6+x)$
$=(2x^2-x-6)(2x^2+x-6)$
l/ $9x^4-10x^2+1$
$=(3x^2)^2-2.1.3x^2+1-4x^2$
$=(3x^2-1)^2-(2x)^2$
$=(3x^2-2x-1)(3x^2+2x-1)$
m/ $x^4-2x^3-7x^2+8x+12$
$=x^4+x^3-3x^3-3x^2-4x^2-4x+12x+12$
$=x^3(x+1)-3x^2(x+1)-4x(x+1)+12(x+1)$
$=(x+1)(x^3-3x^2-4x+12)$
$=(x+1)[x^2(x-3)-4(x-3)]$
$=(x+1)(x-3)(x^2-4)$
$=(x+1)(x-3)(x-2)(x+2)$
