Phân tích đa thức thành nhân tử:
3x^2+7x-6
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\(=2x^4+6x^3-3x^3-9x^2-3x^2-9x+2x+6\)
\(=2x^3\left(x+3\right)-3x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(2x^3-4x^2+x^2-2x-x+2\right)=\left(x+3\right)\left(x-2\right)\left(2x^2+x-1\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(2x^2+2x-x-1\right)=\left(x+3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)\)
2x^4+3x^3-12x^2-7x+6 = (2x^4-x^3)+(4x^3-2x^2)-(10x^2-5x)-(12x-6)
= x^3.(2x-1)+2x^2.(2x-1)-5x.(2x-1)-6.(2x-1) = (2x-1).(x^3+2x^2-5x-6)
= (2x-1).[ (x^3+x^2)+(x^2+x)-(6x+6) ] = (2x-1).(x+1).(x^2+x-6) = (2x-1).(x-1).[(x^2-2x)+(3x-6)]
= (2x-1).(x+1).(x-2).(x+3)
k mk nha
Ta có:x(3x2+4x-7)=x[(3x2-3x)+(7x-7)]=x[3x(x-1)+7(x-1)]=x(x-1)(3x+7)
a) x3 - 7x - 6 = x3 + x2 - x2 - x - 6x - 6
= x2(x + 1) - x(x + 1) - 6(x + 1)
= (x + 1)(x2 - x - 6)
= (x + 1)(x2 + 2x - 3x - 6)
= (x + 1)[x(x + 2) - 3(x + 2)]
= (x + 1)(x + 2)(x - 3)
Dễ mà ^_^: 3x2-7x+2=3x2 -x-6x+2=(3x2-x)-(6x-2)=x(3x-1)-2(3x-1)=(3x-1)(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)3x^2-7x+2
=3x^2-x-6x+2
=(3x^2-x)-(6x-2)
=x(3x-1)-2(3x-1)
=(x-2)(3x-1)
cho loi nx nha may ban
\(3x^2+7x-6\)
\(=3x^2+9x-2x-6\)
\(=3x\left(x+3\right)-2\left(x+3\right)\)
\(=\left(3x-2\right)\left(x+3\right)\)