Violympic toán 8

TK

phân tích các đa thức sau thành nhân tử :

a, x\(^2\)+9x+20

b, x\(^2\)+x-12

c, 6x\(^2\)-11x-16

d, 4x\(^2\)-8x-5

e, x\(^3\)-6x\(^2\)-x+30

g, x\(^3\)+9x\(^2\)+23x+15

h, 2x\(^4\)-x\(^3\)-9x\(^2\)+13x

i, x\(^4\)+2x\(^3\)-16x\(^2\)-2x+15

NT
19 tháng 7 2020 lúc 20:42

a) Ta có: \(x^2+9x+20\)

\(=x^2+4x+5x+20\)

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

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

b) Ta có: \(x^2+x-12\)

\(=x^2+4x-3x-12\)

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

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

c) Ta có: \(6x^2-11x-16\)

\(=6\left(x^2-\frac{11}{6}x-\frac{16}{6}\right)\)

\(=6\left(x^2-2\cdot x\cdot\frac{11}{12}+\frac{121}{144}-\frac{505}{144}\right)\)

\(=6\left[\left(x-\frac{11}{12}\right)^2-\frac{505}{144}\right]\)

\(=6\left(x-\frac{11+\sqrt{505}}{12}\right)\left(x-\frac{11-\sqrt{505}}{12}\right)\)

d) Ta có: \(4x^2-8x-5\)

\(=4x^2-10x+2x-5\)

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

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

e) Ta có: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

\(=\left(x+2\right)\left(x^2-3x-5x+15\right)\)

\(=\left(x+2\right)\left[x\left(x-3\right)-5\left(x-3\right)\right]\)

\(=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)

g) Ta có: \(x^3+9x^2+23x+15\)

\(=x^3+x^2+8x^2+8x+15x+15\)

\(=x^2\left(x+1\right)+8x\left(x+1\right)+15\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+8x+15\right)\)

\(=\left(x+1\right)\left(x^2+3x+5x+15\right)\)

\(=\left(x+1\right)\left[x\left(x+3\right)+5\left(x+3\right)\right]\)

\(=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

h) Ta có: \(2x^4-x^3-9x^2+13x\)

\(=x\left(2x^3-x^2-9x+13\right)\)

i) Ta có: \(x^4+2x^3-16x^2-2x+15\)

\(=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\)

\(=x^3\left(x-3\right)+5x^2\left(x-3\right)-x\left(x-3\right)-5\left(x-3\right)\)

\(=\left(x-3\right)\left(x^3+5x^2-x-5\right)\)

\(=\left(x-3\right)\left[x^2\left(x+5\right)-\left(x+5\right)\right]\)

\(=\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\)

\(=\left(x-3\right)\left(x+5\right)\left(x-1\right)\left(x+1\right)\)

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