Question

Phân tích đa thức thành nhân tử

a/x^3+3xy+y^3-1

b/ 3x^2+22xy+11x+37y+7y^2+10

c/ x^2+2xy+y^2-x-y-12

d/x^8-8x+63

e/x^4-8x+63

g/64x^4+y^4

h/a^6+a^4+a^2b^2+b^4-b^6

Answer 1

a)\(x^3+3xy+y^3-1\)

\(=x^3+3x^2y+3xy^2+y^3-1-3x^2y-3xy^2+3xy\)

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

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

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

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

b) Đặt \(B=3x^2+22xy+11x+37y+7y^2+10\)

Giả sử \(B=\left(ax+by+c\right)\left(mx+ny+p\right)\)

\(=amx^2+anxy+apx+bmxy+bny^2+bpy+cmx+cny+cp\)

\(=amx^2+\left(an+bm\right)xy+\left(ap+cm\right)x+bny^2+\left(bp+cn\right)y+cp\)

Ta được hệ: \(\left\{{}\begin{matrix}am=3;an+bm=22\\ap+cm=11;bn=7\\bp+cn=37;cp=10\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=3;b=1\\c=5;m=1\\n=7;p=2\end{matrix}\right.\)

Vậy B phân tích được thành \(\left(3x+y+5\right)\left(x+7y+2\right)\).