Question

Phân tích đa thức thành nhân tử ( phương pháp thên bớt )

a/ x⁴ + 5x³ + 10x -4

b/ x³ + y³ + z³ - 3xyz

c/ x⁷ + x² + 1

d/ x⁸ + x + 1

e/ x⁵ + x⁴ + 1

g/ x¹⁰ + x⁵ + 1

Answer 1

a)x⁴+5x³+10x-4

=x⁴+2x²+5x³+10x-2x²-4

=x²(x²+2)+5x(x²+2)-2(x²+2)

=(x²+5x-2)(x²+2)

Answer 2

b)x³+y³+z³-3xyz

= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).

Answer 3

c)x⁷+x²+1

=x⁷-x+x²+x+1

=x(x⁶-1)+x²+x+1

=x(x³-1)(x³+1)+x²+x+1

=x(x-1)(x²+x+1)(x³+1)+x²+x+1

=(x²+x+1)(x⁵-x⁴+x²-x)

Answer 4

e)x⁵+x⁴+1

=x⁵-x²+x⁴-x+x²+x+1

=x²(x³-1)+x(x³-1)+x²+x+1

=(x²+x)(x³-1)+x²+x+1

=(x²+x)(x-1)(x²+x+1)+x²+x+1

=(x²+x+1)(x³-x+1)

Answer 5

g)x¹⁰+x⁵+1

=x¹⁰-x+x⁵-x²+x²+x+1

=x(x⁹-1)+x²(x³-1)+x²+x+1

=x(x³-1)(x⁶+x³+1)+x²(x-1)(x²+x+1)+x²+x+1

=(x⁷+x⁴+x)(x-1)(x²+x+1)+x²(x-1)(x²+x+1)+x²+x+1

=(x²+x+1)(x⁸-x⁷+x⁵-x⁴+2x²-x+1)

Answer 6

d)x⁸+x+1

=x⁸-x²+x²+x+1

=x²(x⁶-1)+x²+x+1

=x²(x³-1)(x³+1)+x²+x+1

=(x⁵+x²)(x-1)(x²+x+1)+x²+x+1

=(x²+x+1)(x⁶-x⁵+x³-x²+1)

Answer 7

a) x⁴ + 5x³ + 10x - 4

= (x⁴ - 4) + (5x³ + 10x)

= (x² + 2)(x² - 2) + 5x(x² + 2)

= (x² + 2)(x² - 2 + 5x)

b) x³ + y³ + z³ - 3xyz

= x³ + y³ + z³ + 3x²y + 3xy² - 3x²y - 3xy² - 3xyz

= (x³ + 3x²y + 3xy² + y³) + z³ - (3x²y + 3xy² + 3xyz)

= (x + y)³ + z³ - 3xy(x + y + z)

= (x + y + z)[(x + y)² - (x + y)z + z²] - 3xy(x + y + z)

= (x + y + z)[(x + y)² - (x + y)z + z² - 3xy]

= (x + y + z)(x² + 2xy + y² - xz - yz + z² - 3xy)

= (x + y + z)(x² + y² + z² - xz - yz - xy)

c) x⁷ + x² + 1

= x⁷ + x² + 1 - x + x

= (x⁷ - x) + (x² + x + 1)

= x(x⁶ - 1) + (x² + x + 1)

= x(x³ - 1)(x³ + 1) + (x² + x + 1)

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

= (x² + x + 1)[x(x³ + 1)(x - 1) + 1]

= (x² + x + 1)[(x⁴ + x)(x - 1) + 1]

= (x² + x + 1)(x⁵ - x⁴ + x² - x + 1)

d) x⁸ + x + 1

= x⁸ + x + 1 + x² - x²

= (x⁸ - x²) + (x² + x + 1)

= x²(x⁶ - 1) + (x² + x + 1)

= x²(x³ + 1)(x³ - 1) + (x² + x + 1)

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

= (x² + x + 1)[x²(x³ + 1)(x - 1) + 1]

= (x² + x + 1)[(x⁵ + x²)(x - 1) + 1]

= (x² + x + 1)(x⁶ - x⁵ + x³ - x² + 1)

e) x⁵ + x⁴ + 1

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

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

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

= (x³ - x + 1)(x² + x + 1)