a/ x^4 + x^3 + 2x^2 + x + 1

= (x^4 + 2x^2 + 1) + (x^3 + x)

= (x^2 + 1)^2 + x (x^2 + 1)

= (x^2 + 1) (x^2 + 1 + x)

= (x^2 + 1) (x + 1)^2

Câu 1:Phân tích

a)\(x^2-6x+8\)

b)\(4x^2-7x+3\)

c)\(\left(3x-1\right)^2-\left(2x-3\right)^2\)

d)\(5x-3x-2\)

b, \(4x^2-7x+3=4x^2-4x-3x+3\)

\(=4x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(4x-3\right)\left(x-1\right)\)

c, \(\left(3x-1\right)^2-\left(2x-3\right)^2\)

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

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

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

8x^2+ 4xy- 2ax-ay

\(8x^2+4xy-2ax-ay\)

= \(\left(8x^2+4xy\right)-\left(2ax+ay\right)\)

= \(4x\left(2x+y\right)-a\left(2x+y\right)\)

= \(\left(2x+y\right)\left(4x-a\right)\)

2x^2 - 2xy - 5x + 5y

\(x^4+2x^3+3x^2+2x+1\)

\(=x^4+x^3+x^3+x^2+x^2+x^2+x+x+1\)

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

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

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

phân tích đa thức thành nhân tử: \(x^{16}+x^8-2\)

Phân tích đa thức thành nhân tử

a) ab(a2-b2)+bc(b2-c2)+ca(c2-a2)

phân tích da thức bằng phương pháp hệ sô bất định

x^4+2x^3+3x^2+2x+1

\(ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)\)

\(ab\left[\left(a+b\right)\left(a-b\right)\right]+bc\left[\left(b+c\right)\left(b-c\right)\right]+ca\left[\left(c+a\right)\left(c-a\right)\right]\)

Mình nghĩ như vậy ko bt đúng hay ko

\(x^{16}+x^8-2\)

\(=x^{16}+2x^8-x^8-2\)

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

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

\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+2\right)\)

\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)

2\(x^2\) - 2xy - 5x + 5y

= 2x(x - y) -5(x - y)

=(x - y)(2x - 5)