Phân tích các đa thức sau thành nhân tử

a) x² + 4xy – 21y²

b) 5x² + 6xy + y²

c) x² + 2xy – 15y²

d) (x – y)² + 4(x – y) – 12

e) x² – 7xy + 10y²

f) x²yz + 5xyz – 14yz

g) x⁴ + 4x² – 5

h) x³ – 19x – 30

i) x³ – 5x² – 14x

j) x³ – 7x – 6

k) x³ – 5x² – 14

a) x² + 4xy – 21y²

= x² + 7xy – 3xy – 21y²

= x(x + 7y) – 3y(x + 7y)

= (x + 7y)(x – 3y)

b) 5x² + 6xy + y²

= 5x² + 5xy + xy + y²

= 5x(x + y) + y(x + y)

= (x + y)(5x + y)

c) x² + 2xy – 15y²

= x² + 5xy – 3xy – 15y²

= x(x + 5y) – 3y(x + 5y)

= (x + 5y)(x – 3y)

d) (x – y)² + 4(x – y) – 12

= (x – y)² + 6 (x – y) – 2(x – y) – 12

= (x – y) (x – y + 6) – 2 (x – y + 6)

= (x – y + 6)(x – y – 2)

e) x² – 7xy + 10y²

= x² – 2xy – 5xy + 10y²

= x(x – 2y) – 5y(x – 2y)

= (x – 2y)(x – 5y)

 f) x²yz + 5xyz – 14yz

= x²yz + 7xyz – 2xyz – 14yz

= xyz (x + 7) – 2yz(x + 7)

= yz(x + 7)(x – 2)

g) x⁴ + 4x² – 5

= x⁴ – x² + 5x² – 5

= x² (x² – 1) + 5 (x² – 1)

= (x² – 1) (x² + 5)

= (x – 1)(x + 1)(x² + 5)

h) x³ – 19x – 30

= x³ + 5x² + 6x – 5x² – 25x – 30

= x (x² + 5x + 6) – 5 (x² + 5x + 6)

= (x² + 5x + 6) (x – 5)

= (x – 5)(x² + 2x + 3x + 6)

= (x – 5)[x(x + 2) + 3(x + 2)]

= (x – 5)(x + 2)(x + 3).

i) x³ – 5x² – 14x

= x(x² – 5x – 14)

= x(x² – 7x + 2x – 14)

= x[x(x – 7) + 2 (x – 7)]

= x(x – 7)(x + 2)

j) x³ – 7x – 6

= x³ + 3x² + 2x – 3x² – 9x – 6

= x(x² + 3x + 2) – 3(x² + 3x + 2)

= (x – 3)(x² + 3x + 2)

= (x – 3)(x² + x + 2x + 2)

= (x – 3)[x(x + 1) + 2(x + 1)]

= (x – 3)(x + 1)(x + 2).