\(x^3+9x^2+26x+24\)

\(=x^3+3x^2+6x^2+18x+8x+24\)

\(=\left(x^3+3x^2\right)+\left(6x^2+18x\right)+\left(8x+24\right)\)

\(=x^2\left(x+3\right)+6x\left(x+3\right)+8\left(x+3\right)\)

\(=\left(x+3\right)\left(x^2+6x+8\right)\)

\(=\left(x+3\right)\left(x^2+2x+4x+8\right)\)

\(=\left(x+3\right)\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\)

\(=\left(x+3\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)

\(=\left(x+3\right)\left(x+2\right)\left(x+4\right)\)

\(15^3+29x^2-8x-12=15x^3+30x^2-x^2-2x-6x-12\)

= \(15x^2.\left(x+2\right)-x.\left(x+2\right)-6.\left(x+2\right)\)= \(\left(x+2\right).\left(15x^2-x-6\right)\)

= \(\left(x+2\right).\left(15x^2-10x+9x-6\right)\)= \(\left(x+2\right).\left(3x-2\right).\left(5x+3\right)\)

\(x^3+9x^2+26x+24=x^3+3x^2+6x^2+18x+8x+24\)\(=x.^2\left(x+3\right)+6x.\left(x+3\right)+8.\left(x+3\right)\)\(=\left(x+3\right).\left(x^2+6x+8\right)\)\(\left(x+3\right).\left(x^2+2x+4x+8\right)=\left(x+2\right).\left(x+3\right).\left(x+4\right)\)

Ta có : 15x³ + 29x² - 8x - 12

= 15x³ + 30x² - x² - 8x - 12

= 15x(x + 2) - (8x + 16) - (x² - 4)

= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)

= (x + 2)(15x - 8 - x + 2)

= (x + 2) (14x - 6) 

 click zô nha >_<                 

Ta có : 15x³ + 29x² - 8x - 12

= 15x³ + 30x² - x² - 8x - 12

= 15x(x + 2) - (8x + 16) - (x² - 4)

= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)

= (x + 2)(15x - 8 - x + 2)

= (x + 2) (14x - 6)

(x+2)(14x-6)

a,x^4+2x^3-4x-4

=(x^3+2x^3)-(4x+4)

=x^3(x+2)-4(x+2)

=(x^3-4)(x+2)

\(X^4+2X^3-4X-4\)

\(=\left(X^2\right)^2+2X^3-4X-2^2\)

\(=\left[\left(X^2\right)^2-2^2\right]+\left[2X^3-4X\right]\)

\(=\left(X^2+2\right)\left(X^2-2\right)+2X\left(X^2-2\right)\)

\(=\left(X^2-2\right)\left(X^2+2+2X\right)\)

Nếu bạn muốn có lời giải thì ít thôi @.@

Katherine Lilly Filbert nói rất đúng câu hỏi nhiều như vậy ai mà trả lời đc hết cơ chứ

\(x^3+9x^2+26x+24=\left(x^2+7x+12\right)\left(x+2\right)=\left(x+3\right)\left(x+4\right)\left(x+2\right)\)

Ta có: \(x^3+9x^2+26x+24\)

\(=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)\)

\(=x^2\left(x+2\right)+7x\left(x+2\right)+12\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2+7x+12\right)\)

\(=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]\)

\(=\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)

\(=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

x³ + 9x² + 26x + 24

= x³ + 2x² + 7x + 12x + 14x + 24

= ( x³ + 7x² + 12x ) + ( 2x² + 14x + 24 )

= x( x² + 7x + 12 ) + 2( x² + 7x + 12 )

= ( x² + 7x + 12 )( x + 2 )

= ( x² + 3x + 4x + 12 )( x + 2 )

= [ x( x + 3 ) + 4( x + 3 ) ]( x + 2 )

= ( x + 2 )( x + 3 )( x + 4 )

Lời giải:
a.

$x^4+10x^3+26x^2+10x+1$

$=(x^4+10x^3+25x^2)+x^2+10x+1$

$=(x^2+5x)^2+2(x^2+5x)+1-x^2$

$=(x^2+5x+1)^2-x^2=(x^2+5x+1-x)(x^2+5x+1+x)$

$=(x^2+4x+1)(x^2+6x+1)$

b.

$x^4+x^3-4x^2+x+1$

$=(x^4-x^2)+(x^3-x^2)+(x-x^2)+(1-x^2)$

$=x^2(x-1)(x+1)+x^2(x-1)-x(x-1)-(x-1)(x+1)$

$=(x-1)[x^2(x+1)+x^2-x-(x+1)]$

$=(x-1)(x^3+2x^2-2x-1)$

$=(x-1)[(x^3-1)+(2x^2-2x)]=(x-1)[(x-1)(x^2+x+1)+2x(x-1)]$

$=(x-1)(x-1)(x^2+x+1+2x)=(x-1)^2(x^2+3x+1)$