Question

Bài 1: Phân tích đa thức thành nhân tử:

a. \(x^3+x^2-x+2\)

b. \(x^3-6x^2-x+30\)

c. \(2x^2-5x-12\)

d. \(6x^2-7x-20\)

e. \(\left(2x^2+4\right)^2+9\)

f. \(x^7+x^5+1\)

g. \(x^8+x^4+1\)

h. \(a^6+a^4+a^2b^2+b^4-b^6\)

Answer 1

 

\\(x^3+x^2-x+2\\)

=x³+2x²-x²-2x+x+2

=(x³+2x²)-(x2+2x)+(x+2)

=x²(x+2)-x(x+2)+(x+2)

=(x+2)(x²-x+1)

 

b. \\(x^3-6x^2-x+30\\)

=x³+2x²-8x²-16x+15x+30

=(x³+2x²)-(8x²+16x)+(15x+30x)

=x²(x+2)-8x(x+2)+15(x+2)

=(x+2)(x²-8x+15)

=(x+2)(x²-5x-3x+15)

=(x+2)[(x²-5x)-(3x-15)]

=(x+2)[x(x-5)-3(x-5)]

=(x+2)(x-5)(x-3)

Answer 2

h)\(a^6+a^4+a^2b^2+b^4-b^6\)

\(=\left(a^4+a^2b^2+b^4\right)+\left(a^6-b^6\right)\)

\(=\left(a^4+a^2b^2+b^4\right)+\left[\left(a^2\right)^3-\left(b^2\right)^3\right]\)

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

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

Answer 3

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

c. \(2x^2-5x-12=2x^2-8x+3x-12=\left(2x^2-8x\right)+\left(3x-12\right)=2x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(2x+3\right)\)

d. \(6x^2-7x-20=6x^2+8x-15x-20=\left(6x^2+8x\right)-\left(15x+20\right)=2x\left(3x+4\right)-5\left(3x+4\right)=\left(3x+4\right)\left(2x-5\right)\)

b. \(x^3-6x^2-x+30=\left(x^3-x\right)-\left(6x^2-30\right)=x\left(x^2-1\right)-6\left(x^2-5\right)=\left(x-6\right)\left(x^2-1\right)\left(x^2-5\right)\)