Nhân theo cột dọc :
(x+3)(x2+3x-5)
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\(A\left(x\right)+B\left(x\right)=8x^3-4x-6+2x^3-x^2-3x+6=10x^3-x^2-7x\)
x3 -2x2 + x -1
x
-x+ 5
5x3 -10x2+5x-5
-x4+ 2x3- x2 +x
-x4+7x3-11x2+6x-5
b: A(x)*C(x)
=(x^3-3x^2+5x-3)(x-3)
=x^4-3x^3-3x^3+9x^2+5x^2-15x-3x+9
=x^4-6x^3+14x^2-18x+9
c: B(x):C(x)
\(=\dfrac{-x^3+3x^2-x^2+3x-6x+18-13}{x-3}\)
=-x^3-x-6-13/x-3
d: A(x)-B(x)
=x^3-3x^2+5x-3+x^3-2x^2+3x-5
=2x^3-5x^2+8x-8
a: =3x^3+12x-9-2x^4-8x^2+6x
=-2x^4+3x^3-8x^2+18x-9
b: \(=\dfrac{-2x^4+x^3+6x^3-3x^2}{2x+1}=-x^3+3x^2\)
a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)
c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)
\(=x^4-14x^2-32\)
\(=\left(x^2-16\right)\left(x^2+2\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)
b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)
\(=\left(x^2+5x+1\right)^2\)
x2+3x-5
x
x+3
3x2+9x -15
x3+3x2 -5x
x3+ 6x2+4x-15