Question

1)Phân tích các đa thức sau thành nhân tử:

a. \(4x^3-\dfrac{1}{2}\)

b.\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

c.\(x^4+64\)

d.\(x^2-11x+24\)

Answer 1

a)\(4x^3-\dfrac{1}{2}\)

\(=4\left(x^3-\dfrac{1}{8}\right)=4\left(x^3-\left(\dfrac{1}{2}\right)^3\right)\)

\(=4\left(x-y\right)\left(x^2+\dfrac{x}{2}+\dfrac{1}{4}\right)\)

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

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

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

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

\(=\left(a+b-c\right)\left(a-b+c\right)\left(-a+b+c\right)\left(a+b+c\right)\)

c)\(x^4+64\)

\(=x^4+16x^2+64-16x^2\)

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

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

d)\(x^2-11x+24\)

\(=x^2-8x-3x+24\)

\(=x\left(x-8\right)-3\left(x-8\right)\)

\(=\left(x-3\right)\left(x-8\right)\)