11x^2-5x-34phân tích đa thức sau thành nhân tử
camown m.n ạ
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1: Đa thức này ko phân tích được nha bạn
2: \(x^2+8x+7\)
\(=x^2+x+7x+7\)
\(=x\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(x+7\right)\)
3: \(x^2-6x-16\)
\(=x^2-8x+2x-16\)
\(=x\left(x-8\right)+2\left(x-8\right)\)
\(=\left(x-8\right)\left(x+2\right)\)
4: \(4x^2-8x+3\)
\(=4x^2-2x-6x+3\)
\(=2x\left(2x-1\right)-3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x-3\right)\)
5: \(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
Ta có:\(x^3+9x^2+11x-21\)
\(=x^3-x^2+10x^2-10x+21x-21=x^2\left(x-1\right)+10x\left(x-1\right)+21\left(x-1\right)\)
\(=\left(x^2+10x+21\right)\left(x-1\right)=\left(x^2+3x+7x+21\right)\left(x-1\right)\)
\(=\left[x\left(x+3\right)+7\left(x+3\right)\right]\left(x-1\right)\)
\(=\left(x+3\right)\left(x+7\right)\left(x-1\right)\)
x^3+9x^2+11x-21=x^3-x^2+10x^2-10x+21x-21=(x^3-x^2)+(10x^2-10x)+(21x-21)
=x^2(x-1)+10x(x-1)+21(x-1)=(x-1)(x^2+10x+21)=(x-1)(x^2+3x+7x+21)=(x-1)[(x^2+3x)+(7x+21)]
=(x-1)(x+7)(x+3)
\(x^3+6x^2+11x+6=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
= \(x^4-2x^3-6x^3+12x^2-x^2+2x+6x-12\)
= \(x^3\left(x-2\right)-6x^2\left(x-2\right)-x\left(x-2\right)+6\left(x-2\right)\)
= \(\left(x-2\right)\left(x^3-6x^2-x+6\right)\)
= \(\left(x-2\right)\left(x^2\left(x-6\right)-\left(x-6\right)\right)\)
= \(\left(x-2\right)\left(x-6\right)\left(x-1\right)\left(x+1\right)\)
x4 - 8x3 + 11x2 + 8x - 12
= (x3 - 7x2 + 4x + 12)(x - 1)
= (x3 - 8x + 12)(x + 1)(x - 1)
= (x - 6)(x - 2)(x + 1)(x - 1)
\(11x^2-5x-34=11x^2-22x+17x-34=11x\left(x-2\right)+17\left(x-2\right)=\left(x-2\right)\left(11x+17\right)\)