a) x³ + 4x² - 29x + 24                                                           

= x³ - 3x² + 7x² - 21x - 8x + 24

= x²(x-3) + 7x(x-3) - 8(x-3)

= (x-3)(x²+7x-8)

=(x-3)(x²+8x-x-8)

= (x-3)[(x²+8x)-(x+8)]

= (x-3)[x(x+8)-(x+8)]

= (x-3)(x+8)(x-1)

\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

\(=\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x+4+1\right)^2\)

\(=\left(x^2+5x+5\right)^2\)

(x + 1)(x + 2)(x + 3)(x + 4) - 24

= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 24

= (x² + 4x + x +4)(x² + 3x + 2x + 12) - 24

= (x² + 5x + 4)(x² + 5x + 12) - 24

Đặt t = x² + 5x + 8

Ta có: x² + 5x + 4 = x² + 5x + 8 - 4 (1)

x² + 5x + 12 = x² + 5x + 8 + 4 (2)

Thay t = x² + 5x + 8 vào (1) và (2), ta có:

⇒ (t - 4)(t + 4) - 24

= t² - 16 - 24

= t² - 40

= (t - \(\sqrt{40}\))(t + \(\sqrt{40}\))

= (x² + 5x + 8 - \(\sqrt{40}\))(x² + 5x + 8 + \(\sqrt{40}\))

= 3²-4\(^2\)(x-1)\(^2\)

=[3-4(x-1)].[(3+4(x-1)]

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

=(7-4x)(4x-1)

a) \(x^2+7x+12\)

\(=x^2+3x+4x+12\)

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

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

b) \(3x^2-5x+2\)

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

\(=3x\left(x-1\right)-2\left(x-1\right)\)

\(=\left(x-1\right)\left(3x-2\right)\)

a) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 4)(x + 3)

b) 3x² - 5x + 2 = 3x² - 3x - 2x + 2 = 3x(x - 1) - 2(x - 1) = (3x - 2)(x - 1)

c) x² + 9x - 10 = x² + 10x - x - 10 = x(x + 10) - (x + 10) = (x - 1)(x + 10)

d) x² - 7x - 8 = x² - 8x + x - 8 = x(x - 8) + (x - 8) = (x + 1)(x - 8)

e) 2x² + 3x - 5 = 2x² + 5x - 2x - 5 = x(2x + 5) - (2x + 5) = (x  - 1)(2x + 5)

a, \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=\left(x+2\right)3x\)

\(-\left(x+2\right)+3\left(x^2-4\right)\)

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

\(=\left(x+2\right)\left[3\left(x-2\right)-1\right]=\left(x+2\right)\left(3x-7\right)\)