Question

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1.phân tích thành nhân tử:

a.45 +x\(^3\)\(-\) 5x\(^2\) \(-\) 9x

b.x^4\(-\)2x^3\(-\)2x^2\(-\)2x\(-\)3

c.2x^2+3x\(-\)5

d.x^4+4y^4

e.(x+2)(x+4)(x+6)(x+8)+16

Answer 1

[Trần Việt Linh]

a) \(45+x^3-5x^2-9x\)

\(=\left(x^3-5x^2\right)-\left(9x-45\right)\)

\(=x^2\left(x-5\right)-9\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2-9\right)=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)

Answer 2

[Nguyễn Hải Anh Jmg]

\(a,45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)+\left(45-9x\right)\)
\(=x^2\left(x-5\right)+9\left(5-x\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)\)
\(=\left(x-5\right)\left(x^2-3^2\right)\)
\(=\left(x-5\right)\left(x+3\right)\left(x-3\right)\)
\(c,2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
\(e,\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\left(2\right)\)
(1)\(\Leftrightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\left(3\right)\)
Thay (3) vào (2),ta được:\(\left(x^2+10x+20\right)^2\)