Phân tích đa thức sau thành nhân tử:
9x^4+15x^3+43x^2+22x-40
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CN
2
9 tháng 9 2015
\(x^3+x^2+9x-10x^2-10x+25x+25\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+25\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+25\right)=\left(x+1\right)\left(x-5\right)^2\)
28 tháng 11 2016
p/ x4 - 9x3 + 22x2 - 9x + 1
= (x4 - 5x3 + x2) + (- 4x3 + 20x2 - 4x) + (x2 - 5x + 1)
= (x2 - 5x + 1)(x2 - 4x + 1)
q) x4 - 6x3 + 14x2 - 22x + 5
= (x4 - 4x3 + x2) + (- 2x3 + 8x2 - 2x) + (5x2 - 20x + 5)
= (x2 - 4x + 1)(x2 - 2x + 5)
HT
2
2 tháng 11 2016
\(2x^4+x^3-22x^2+15x-36\)
\(=\left(2x^4-6x^3\right)+\left(7x^3-21x^2\right)+\left(-x^2+3x\right)+\left(12x-36\right)\)
\(=\left(x-3\right)\left(2x^3+7x^2-x+12\right)\)
\(=\left(x-3\right)\left(\left(2x^3+8x^2\right)+\left(-x^2-4x\right)+\left(3x+12\right)\right)\)
\(=\left(x-3\right)\left(x+4\right)\left(2x^2-x+3\right)\)
KT
31 tháng 7 2018
mk ghi kết quả thôi nhé, nếu từ kết quả mak k biết biến đổi thì ib cho mk
\(x^5-7x^4-x^3+43x^2-36=\left(x-6\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
câu thứ 2 bạn ktra lại đề
\(x^4+2x^3-15x^2-18x+64=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
\(x^3-x^2-4=\left(x-2\right)\left(x^2+x+2\right)\)
\(x^3-3x^2-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
KT
1 tháng 8 2018
a) \(x^5-7x^4-x^3+43x^2-36\)
\(=x^3\left(x^2-1\right)-7x^2\left(x^2-1\right)+36\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^3-7x^2+36\right)=\left(x-1\right)\left(x+1\right)\left(x^3+2x^2-9x^2-18x+18x+36\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x^9-9x+18\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x-6\right)\)
c) \(x^4+2x^3-15x^2-18x+64\)
\(=x^3\left(x-2\right)+4x^2\left(x-2\right)-7x\left(x-2\right)-32\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
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
\(15^3+29x^2-8x-12=15x^3+30x^2-x^2-2x-6x-12\)
= \(15x^2.\left(x+2\right)-x.\left(x+2\right)-6.\left(x+2\right)\)= \(\left(x+2\right).\left(15x^2-x-6\right)\)
= \(\left(x+2\right).\left(15x^2-10x+9x-6\right)\)= \(\left(x+2\right).\left(3x-2\right).\left(5x+3\right)\)
\(x^3+9x^2+26x+24=x^3+3x^2+6x^2+18x+8x+24\)\(=x.^2\left(x+3\right)+6x.\left(x+3\right)+8.\left(x+3\right)\)\(=\left(x+3\right).\left(x^2+6x+8\right)\)\(\left(x+3\right).\left(x^2+2x+4x+8\right)=\left(x+2\right).\left(x+3\right).\left(x+4\right)\)
=9x^4-6x^3+21x^3-14x^2+57x^2-38x+60x-40
=(3x-2)(3x^3+7x^2+19x+20)
=(3x-2)(3x+4)(x^2+x+5)
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