Phân tích thành nhân tử: x^3-5x+4
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#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
\(x^3+5x^2+8x-4=x^3+x^2+4x^2+4x+4x+4\)
\(=\left(x^3+x^2\right)+\left(4x^2+4x\right)+\left(4x+4\right)\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x^2+4x+4\right)\left(x+1\right)\)
\(=\left(x+2\right)^2\left(x+1\right)\)
Dễ mà:
f(x)=(2x4-2x3)-(3x3-3x2)-(8x2-8x)-(3x-3)
=2x3(x-1)-3x2(x-1)-8x(x-1)-3(x-1)
=(x-1)(2x3-3x2-8x-3)
=(x-1)[(2x3+2x2)-(5x2+5x)-(3x+3)]
=(x-1)[2x2(x+1)-5x(x+1)-3(x+1)]
=(x-1)(x+1)(2x2-5x-3)
=(x-1)(x+1)[2x(x-3)+(x-3)]
=(x-1)(x+1)(x-3)(2x+1)
\(f\left(x\right)=2x^4-5x^3-5x^2+5x+3.\)
\(=\left(2x^4-2x^3\right)-\left(3x^3-3x^2\right)-\left(8x^2-8x\right)-\left(3x-3\right)\text{ }\left(\text{Hơi khó hiểu thông cảm! }\right)\)
\(=2x^3\left(x-1\right)-3x^2\left(x-1\right)-8x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(2x^3-3x^2-8x-3\right)\)
\(=\left(x-1\right)\left[\left(2x^3+2x^2\right)-\left(5x^2+5x\right)-\left(3x+3\right)\right]\)
\(=\left(x-1\right)\left[2x^2\left(x+1\right)-5x\left(x+1\right)-3\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(2x^2-5x-3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left[\left(2x^2-6x\right)+\left(x-3\right)\right]\)
\(=\left(x-1\right)\left(x+1\right)\left(x-3\right)\left(2x+1\right)\)
\(x^3-5x^2+8x-4\)
\(=x^3-4x^2-x^2+4x+4x-4\)
\(=\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)\)
\(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
Xong rùi đấy
\(a,x^4+5x^3-8x-40=x^3\left(x+5\right)-8\left(x+5\right)\\ =\left(x^3-8\right)\left(x+5\right)=\left(x-2\right)\left(x^2+2x+4\right)\left(x+5\right)\\ b,3x^2-6x-12y^2+3=3\left(x^2-2x-4y^2+1\right)\\ =3\left[\left(x-1\right)^2-4y^2\right]=3\left(x-2y-1\right)\left(x+2y-1\right)\)