phân tích đa thức thành nhân tử: \(5x^2-26x+24\)
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NV
3
A
14 tháng 10 2020
\(x^3+9x^2+26x+24=\left(x^2+7x+12\right)\left(x+2\right)=\left(x+3\right)\left(x+4\right)\left(x+2\right)\)
14 tháng 10 2020
Ta có: \(x^3+9x^2+26x+24\)
\(=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)\)
\(=x^2\left(x+2\right)+7x\left(x+2\right)+12\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+7x+12\right)\)
\(=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]\)
\(=\left(x+2\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)
\(=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
9 tháng 1 2022
\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24\\ =\left[\left(x^2+5x\right)^2-6\left(x^2+5x\right)\right]+\left[4\left(x^2+5x\right)-24\right]\\ =\left(x^2+5x\right)\left(x^2+5x-6\right)+4\left(x^2+5x-6\right)\\ =\left(x^2+5x-6\right)\left(x^2+5x+4\right)\\ =\left(x^2-x+6x-6\right)\left(x^2+4x+x+4\right)\\ =\left[x\left(x-1\right)+6\left(x-1\right)\right]+\left[x\left(x+4\right)+\left(x+4\right)\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4\right)\left(x+6\right)\)
NH
6 tháng 7 2017
\(x^3+9x^2+26x+24\)
\(=x^3+3x^2+6x^2+18x+8x+24\)
\(=\left(x^3+3x^2\right)+\left(6x^2+18x\right)+\left(8x+24\right)\)
\(=x^2\left(x+3\right)+6x\left(x+3\right)+8\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+6x+8\right)\)
\(=\left(x+3\right)\left(x^2+2x+4x+8\right)\)
\(=\left(x+3\right)\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\)
\(=\left(x+3\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]\)
\(=\left(x+3\right)\left(x+2\right)\left(x+4\right)\)
KC
6 tháng 7 2017
\(15^3+29x^2-8x-12=15x^3+30x^2-x^2-2x-6x-12\)
= \(15x^2.\left(x+2\right)-x.\left(x+2\right)-6.\left(x+2\right)\)= \(\left(x+2\right).\left(15x^2-x-6\right)\)
= \(\left(x+2\right).\left(15x^2-10x+9x-6\right)\)= \(\left(x+2\right).\left(3x-2\right).\left(5x+3\right)\)
\(x^3+9x^2+26x+24=x^3+3x^2+6x^2+18x+8x+24\)\(=x.^2\left(x+3\right)+6x.\left(x+3\right)+8.\left(x+3\right)\)\(=\left(x+3\right).\left(x^2+6x+8\right)\)\(\left(x+3\right).\left(x^2+2x+4x+8\right)=\left(x+2\right).\left(x+3\right).\left(x+4\right)\)
6 tháng 7 2017
Ta có : 15x3 + 29x2 - 8x - 12
= 15x3 + 30x2 - x2 - 8x - 12
= 15x(x + 2) - (8x + 16) - (x2 - 4)
= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)
= (x + 2)(15x - 8 - x + 2)
= (x + 2) (14x - 6)
click zô nha >_<
NT
0
6 tháng 7 2017
Ta có : 15x3 + 29x2 - 8x - 12
= 15x3 + 30x2 - x2 - 8x - 12
= 15x(x + 2) - (8x + 16) - (x2 - 4)
= 15x(x + 2) - 8(x + 2) - (x - 2)(x + 2)
= (x + 2)(15x - 8 - x + 2)
= (x + 2) (14x - 6)
NQ
0
=5x^2-6x-20x+24=x(5x-6)+4(5x-6)=(5x-6)(x-4)