Question

Phân tích đa thức thành nhân tử

a.\(a^4+a^2+1\)

b.\(a^4+a^2-2\)

c.\(x^4+4x^2-5\)

d.\(x^3-10x-12\)

e.\(6x^3-17x^2+14x-3\)

g.\(4x^3-25x^2-53x-24\)

h.\(x^4-34x^2+225\)

i.\(x^3-5x^2y-14xy^2\)

k.\(4x^4-12x^2+1\)

l.\(2x^4+5x^3+13x^2+25x+15\)

m.\(x^3-19x+30\)

n.\(x^3+9x^2+26x+24\)

o.\(x^4+3x^3+x^2-12x-20\)

Answer 1

a)\(a^4+a^2+1=\left(a^2\right)^2+2a^2.1+1^2-a^2=\left(a^2+1\right)^2-a^2=\left(a^2+1+a\right)\left(a^2+1-a\right)\)

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

c)\(x^4+4x^2-5=x^4-x^2+5x^2-5=x^2\left(x^2-1\right)+5\left(x^2-1\right)=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)

d)\(\left(x+2\right)\left(x^2-2x-6\right)=x^3-2x^2-6x+2x^2-4x-12=x^3-10x-12\)

\(\Rightarrow x^3-10x-12=\left(x+2\right)\left(x^2-2x-6\right)\)

e)\(6x^3-17x^2+14x-3\)

Ta có: \(\left(ax^2+bx+c\right)\left(dx+e\right)\)

\(=adx^3+aex^2+bdx^2+bex+cdx+ce\)

\(=adx^3+\left(ae+bd\right)x^2+\left(be+cd\right)x+ce\)

Do đó:\(\left\{{}\begin{matrix}ad=6\\ae+bd=-17\\be+cd=14\\ce=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3;b=-4\\c=1;d=2\\e=-3\end{matrix}\right.\)

Suy ra: \(6x^3-17x^2+14x-3=\left(3x^2-4x+1\right)\left(2x-3\right)\)

Answer 2

h)\(x^4-34x^2+225=x^4-15x^2-15x^2+225-4x^2=x^2\left(x^2-15\right)-15\left(x^2-15\right)-\left(2x\right)^2=\left(x^2-15\right)^2-\left(2x\right)^2=\left(x^2+2x-15\right)\left(x^2-2x-15\right)=\left(x^2-3x+5x-15\right)\left(x^2+5x-3x-15\right)=\left[\left(x-3\right)\left(x+5\right)\right]^2\)

Answer 3

i)\(x^3-5x^2y-14xy^2=x^3+2x^2y-7x^2y-14xy^2=x^2\left(x+2y\right)-7xy\left(x+2y\right)=\left(x^2-7xy\right)\left(x+2y\right)\)