a: \(x^3-9x^2+6x+16\)

\(=x^3-8x^2-x^2+8x-2x+16\)

\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)

\(=\left(x-8\right)\left(x^2-x-2\right)\)

\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)

b: \(x^3-x^2-x-2\)

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

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

\(=\left(x-2\right)\cdot\left(x^2+x+1\right)\)

c: \(x^3+x^2-x+2\)

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

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

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

d: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

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

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

\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)

e: Sửa đề: \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

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

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

f: \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

g: \(2x^3-x^2+5x+3\)

\(=2x^3+x^2-2x^2-x+6x+3\)

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

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

h: \(\left(x^2-3\right)^2+16\)

\(=x^4-6x^2+9+16\)

\(=x^4-6x^2+25\)

\(=x^4+10x^2+25-16x^2\)

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

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

=> x (x³ + 5x² - 7x - 41)

Thời gian có hạn copy cái này hộ mình vào google xem nha :

https://lazi.vn/quiz/d/16491/nhac-edm-la-loai-nhac-the-loai-gi

Vào xem xong các bạn nhận được 1 thẻ cào mệnh giá 100k nhận thưởng bằng cách nhắn tin vs mình và 1 phần thưởng bí mật là chiếc áo đá bóng,....

Có 300 giải nhanh nha đã có 241 người nhận rồi

OKthanks

\(x^4+5x^3-7x^2-41x-30\)

\(=x^4+x^3+4x^3+4x^2-11x^2-11x-30x-30\)

\(=x^3\left(x+1\right)+4x^2\left(x+1\right)-11x\left(x+1\right)-3x\left(x+1\right)\)

\(=\left(x+1\right)\left(x^3+4x^2-11x-30\right)\)

\(=\left(x+1\right)\left(x^3-3x^2+7x^2-21x+10x-30\right)\)

\(=\left(x+1\right)\left[x^2\left(x-3\right)+7x\left(x-3\right)+10x\left(x-3\right)\right]\)

\(=\left(x+1\right)\left(x-3\right)\left(x^2+2x+5x+10\right)\)

\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\left(x+5\right)\)

1, x³+ 6x²+11x+6

= x³ + 2x² + 4x² + 8x + 3x + 6 

= x²(x + 2) + 4x(x + 2) + 3(x + 2)

= (x + 2)(x² + 4x + 3)

2, x⁴+3x³-7x²-27x-18

= x⁴ + 3x³ - 9x² + 2x² - 27x -18

= (x⁴ - 9x²) + (3x³ - 27x) + (2x² - 18)

= x²(x² - 9) + 3x(x² - 9) + 2(x² - 9)

= (x² - 9)(x² + 3x + 2)

= (x + 3)(x - 3)(x² + 3x + 2)

3, x³-8x²+x+42

= x³ - 7x² - x² + 7x - 6x + 42

= (x³ - 7x²) - (x² - 7x) - (6x - 42)

= x²(x - 7) - x(x - 7) - 6(x - 7)

= (x - 7)(x² - x - 6) 

4, x⁴+5x³-7x²-41x-30 

= x⁴ + x³ + 4x³ - 4x² - 11x² - 11x - 30x - 30

= (x⁴ + x³) + (4x³ - 4x²) - (11x² + 11x) - (30x + 30)

= x³(x + 1) + 4x²(x + 1) - 11x(x + 1) - 30(x + 1)

= (x³ + 4x² - 11x - 30)(x + 1)

5, x⁵+x-1

= x⁵ - x⁴ + x³ + x⁴ - x³ + x² - x²+ x -1 

= x³(x² - x + 1)+ x²(x² - x + 1)- (x² - x + 1) 

= (x² - x + 1)(x³ + x² - 1)

6, x⁵-x⁴-1

= x⁵ - x³ - x² - x⁴ + x² + x + x³ - x - 1 

= x²(x³ - x - 1) - x(x³ - x - 1) + (x³ - x - 1)

= (x² - x + 1)(x³ - x - 1)

1, x 3+ 6x 2+11x+6

= x 3 + 2x 2 + 4x 2 + 8x + 3x + 6

= x 2 ﴾x + 2﴿ + 4x﴾x + 2﴿ + 3﴾x + 2﴿

= ﴾x + 2﴿﴾x 2 + 4x + 3﴿

2, x 4+3x 3‐7x 2‐27x‐18

= x 4 + 3x 3 ‐ 9x 2 + 2x 2 ‐ 27x ‐18

= ﴾x 4 ‐ 9x 2 ﴿ + ﴾3x 3 ‐ 27x﴿ + ﴾2x 2 ‐ 18﴿

= x 2 ﴾x 2 ‐ 9﴿ + 3x﴾x 2 ‐ 9﴿ + 2﴾x 2 ‐ 9﴿

= ﴾x 2 ‐ 9﴿﴾x 2 + 3x + 2﴿

=﴾x + 3﴿﴾x ‐ 3﴿﴾x 2 + 3x + 2﴿

3, x 3‐8x 2+x+42

= x 3 ‐ 7x 2 ‐ x 2 + 7x ‐ 6x + 42

= ﴾x 3 ‐ 7x 2 ﴿ ‐ ﴾x 2 ‐ 7x﴿ ‐ ﴾6x ‐ 42﴿

= x 2 ﴾x ‐ 7﴿ ‐ x﴾x ‐ 7﴿ ‐ 6﴾x ‐ 7﴿

= ﴾x ‐ 7﴿﴾x 2 ‐ x ‐ 6﴿

4, x 4+5x 3‐7x 2‐41x‐30

= x 4 + x 3 + 4x 3 ‐ 4x 2 ‐ 11x 2 ‐ 11x ‐ 30x ‐ 30

= ﴾x 4 + x 3 ﴿ + ﴾4x 3 ‐ 4x 2 ﴿ ‐ ﴾11x 2 + 11x﴿ ‐ ﴾30x + 30﴿

= x 3 ﴾x + 1﴿ + 4x 2 ﴾x + 1﴿ ‐ 11x﴾x + 1﴿ ‐ 30﴾x + 1﴿

= ﴾x 3 + 4x 2 ‐ 11x ‐ 30﴿﴾x + 1﴿

5, x 5+x‐1

= x 5 ‐ x 4 + x 3 + x 4 ‐ x 3 + x 2 ‐ x 2+ x ‐1

= x 3 ﴾x 2 ‐ x + 1﴿+ x 2 ﴾x 2 ‐ x + 1﴿‐ ﴾x 2 ‐ x + 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 + x 2 ‐ 1﴿ 6, x 5‐x 4‐1

= x 5 ‐ x 3 ‐ x 2 ‐ x 4 + x 2 + x + x 3 ‐ x ‐ 1

= x 2 ﴾x 3 ‐ x ‐ 1﴿ ‐ x﴾x 3 ‐ x ‐ 1﴿ + ﴾x 3 ‐ x ‐ 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 ‐ x ‐ 1﴿ 

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

\(x^3-4x^2+8x-8\)

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

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

Câu 1: Đa thức không phân tích được thành nhân tử

Câu 2:

\(x^3-8x^2+x+42\)

\(=x^3+2x^2-10x^2-20x+21x+42\)

\(=x^2(x+2)-10x(x+2)+21(x+2)\)

\(=(x^2-10x+21)(x+2)\)

\(=(x^2-3x-7x+21)(x+2)\)

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

Câu 3:

Đa thức không phân tích được thành nhân tử dù sửa dấu = thành cộng hoặc trừ.

\(=x^2\left(x-6\right)-24\left(x-6\right)=\left(x^2-24\right)\left(x-6\right)\)

\(x^3-6x^2-24x+144=\left(x^3-6x^2\right)-\left(24x-144\right)=x^2\left(x-6\right)-24\left(x-6\right)=\left(x-6\right)\left(x^2-24\right)\)