phân tích đa thức thành nhân tử
4x2-3x-1
x2+3x-10
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= 4x2-y2+8y-16
= 4x2- (y2-8y+16)
= 4x2- (y-4)2
=(4x-y+4) (4x+y-4)
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
`#040911`
`x^2 - 3x - 10`
`= x^2 + 2x - 5x - 10`
`= (x^2 + 2x) - (5x + 10)`
`= x(x + 2) - 5(x + 2)`
`= (x - 5)(x + 2)`
\(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
3x2 - 7x - 10 = 3x2 + 3x - 10x - 10 = 3x(x + 1) - 10(x + 1) = (3x - 10)(x + 1)
a) \(x^3+6x^2+3x-10\)
\(=x^3-x^2+7x^2-7x+10x-10\)
\(=x^2\left(x-1\right)+7x\left(x-1\right)+10\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+7x+10\right)\)
\(=\left(x-1\right)\left(x^2+2x+5x+10\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x+5\right)\)
b) \(x^3+3x^2-33x-35\)
\(=x^3-5x^2+8x^2-40x+7x-35\)
\(=x^2\left(x-5\right)+8x\left(x-5\right)+7\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2+8x+7\right)\)
\(=\left(x-5\right)\left(x^2+x+7x+7\right)\)
\(=\left(x-5\right)\left(x+1\right)\left(x+7\right)\)
Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
\(2x^3-3x^2+3x-1\)
\(=2x^3-x^2-2x^2+x+2x-1\)
\(=x^2\left(2x-1\right)-x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-x+1\right)\)
4x2-3x-1=(3x2-3x)+(x2-1)=3x(x-1)+(x-1)(x+1)=(x-1)(3x+x+1)=(x-1)(4x+1)