Phân tích đa thức thành nhân tử: 8x^3(x-3)+16x^2(3-x)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^4-4x^3+8x^2-16x+16 \)
\(=x^3\left(x-2\right)-2x^2\left(x-2\right)+4x\left(x-2\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[x^2\left(x-2\right)+4\left(x-2\right)\right]\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
Lời giải:
a.
$(xy)^2-xy-2=(x^2y^2+xy)-(2xy+2)$
$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$
b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?
a,x4-4x3+8x2-16x+16
=x4-4x3+4x2+4x2-16x+16
=x2.(x-2)2+4.(x-2)2
=(x-2)2(x2+4)
a) \(4x^4+4x^3-x^2-x=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(4x^3-x\right)\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\)
\(=x\left\{\left(2x\right)^2-1\right\}\left(x+1\right)=x\left(2x-1\right)\left(2x+1\right) \left(x+1\right)\)
c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)=\left(x^2-4x+4\right)\left(x^2+4\right)=\left(x-2\right)^2\left(x^2+4\right)\)
b) \(x^6-x^4-9x^3+9x^2=x^4\left(x^2-1\right)-\left(9x^3-9x^2\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=\left(x^5+x^4-9x^2\right)\left(x-1\right)=\left(x-1\right)x^2\left(x^3+x^2-9\right)\)
x^4 - 4x^3 - 8x^2 - 16x + 16
= x^4-8x^2+16-4x^3-16x
= ( x^2+4)^2 - 4x(x^2+4 )
= ( x^2 + 4 )(x^2 + 4 - 4x)
= (x^2 + 4 )( x - 2 )^2
= ( x3 + 2x2y + xy2 ) - 16x
= x (x2 + 2xy + y2) - 16x
= x( x + y)2 - 16x
= x [ ( x + y)2 - 16 ]
= x ( x + y +4) ( x + y - 4)
a,x4-4x3+8x2-16x+16
=(x4-4x3+4x2)+(4x2-16x+16)
=(x^2-2x)^2+(2x-4)^2
=x^2(x-2)^2+4(x-2)^2
=(x-2)^2(x^2+4)
\(A=x^3-15x^4+16x^3-29x^2\)
\(A=\left(x^3+16x^3\right)-15x^4-29x^2\)
\(A=17x^3-15x^4-29x^2\)
\(A=-15x^4+17x^3-29x^2\)
\(A=-x^2\left(15x^2-17x+29\right)\)
\(=\left(x-3\right)\left(8x^3-16x^2\right)=8x^2\left(x-2\right)\left(x-3\right)\)
\(8x^3\left(x-3\right)+16x^2\left(3-x\right)\)
\(=8x^3\left(x-3\right)-16x^2\left(x-3\right)\)
\(=8x^2\left(x-3\right)\left(x-2\right)\)