Question

Phân tích thành nhân tử

 a) x^4+x^3+6x^2+5x+5 

 b) x^3+x-2

 c) x^3+3x^2-4

 d) xy(x+y) + yz(y+z) + xz(x+z) + 3xyz

Answer 1

a/ \(=x^4+x^3+x^2+5x^2+5x+5\)

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

b/ \(=x^3+x^2+2x-x^2-x-2\)

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

c/ \(=x^3+4x^2+4x-x^2-4x-4\)

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

câu d khó quá , mk lm k nổi , sr nha ^^

Answer 2

a) x⁴ + x³ + 6x² + 5x + 5
= x⁴ + x³ + x² + 5x² + 5x + 5
= x² ( x² + x + 1) + 5 (x² + x + 1)
= (x² + x + 1) (x² + 5)

b) x³ + x - 2
= x³ + x² + 2x - x² - x - 2
= x (x² + x + 2) - (x² + x + 2)
= (x² + x + 2) (x - 1)

c) x³ + 3x² - 4
= x³ + 4x² + 4x - x² - 4x - 4
= x (x² + 4x + 4) - (x² + 4x + 4)
= (x² + 4x + 4) (x - 1)
= (x + 2)² (x - 1)

d) xy(x + y) + yz(y + z) + xz(x + z) + 3xyz
= xy(x + y) + xyz + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y + z) + yz(x + y + z) + xz(x + y + z)
= (x + y + z) (xy + yz + xz)