Phân tích đa thức thành nhân tử:
\(2x^4-10x^3+3x^2-3x+2\)
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10 tháng 9 2021
a: Ta có: \(-3x^4+20x^3-35x^2-10x+48\)
\(=-\left(3x^4-20x^3+35x^2+10x-48\right)\)
\(=-\left(3x^4-9x^3-11x^3+33x^2+2x^2-6x+16x-48\right)\)
\(=-\left(x-3\right)\left(3x^3-11x^2+2x+16\right)\)
\(=-\left(x-3\right)\left(3x^3-6x^2-5x^2+10x-8x+16\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x^2-5x-8\right)\)
\(=-\left(x-3\right)\left(x-2\right)\left(3x-8\right)\left(x+1\right)\)
b: Ta có: \(-\left(2x^4+7x^3+x^2-7x-3\right)\)
\(=-\left(2x^4-2x^3+9x^3-9x^2+10x^2-10x+3x-3\right)\)
\(=-\left(x-1\right)\left(2x^3+9x^2+10x+3\right)\)
\(=-\left(x-1\right)\left(2x^3+2x^2+7x^2+7x+3x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x-1\right)\left(x+1\right)\cdot\left(x+3\right)\left(2x+1\right)\)
DT
1
7 tháng 8 2019
1) =\(-3x^4+9x^3+11x^3-33x^2-2x^2+6x-16x+48\)
=\(-3x^3\left(x-3\right)+11x^2\left(x-3\right)-2x\left(x-3\right)-16\left(x-3\right)\)
= \(\left(x-3\right)\left(-3x^3+11x^2-2x-16\right)\)
= \(\left(x-3\right)\left(-3x^3+6x^2+5x^2-10x+8x-16\right)\)
=\(\left(x-3\right)\left(-3x^2\left(x-2\right)+5x\left(x-2\right)+8\left(x-2\right)\right)\)
= \(\left(x-3\right)\left(x-2\right)\left(-3x^2+5x+8\right)\)
= \(\left(x-3\right)\left(x-2\right)\left(x-\frac{8}{3}\right)\left(x+1\right)\)
Ý b lm theo ý tưởng tương tự nha bn :D
VT
2
1 tháng 12 2021
\(=x^3-3x^2+6x^2-18x+8x-24\\ =\left(x-3\right)\left(x^2+6x+8\right)\\ =\left(x-3\right)\left(x^2+2x+4x+8\right)\\ =\left(x-3\right)\left(x+2\right)\left(x+4\right)\)
1 tháng 12 2021
\(x^3+3x^2-10x-24=\left(x^3-3x^2\right)+\left(6x^2-18x\right)+\left(8x-24\right)=x^2\left(x-3\right)+6x\left(x-3\right)+8\left(x-3\right)=\left(x-3\right)\left(x^2+6x+8\right)=\left(x-3\right)\left[\left(x^2+2x\right)+\left(4x+8\right)\right]=\left(x-3\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]=\left(x-3\right)\left(x+2\right)\left(x+4\right)\)
TQ
1
HN
1 tháng 11 2019
1) \(x^4-2x^3+3x^2-2x+1\)
\(=x^2\left(x^2-x+1\right)-x\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)^2\)
2) \(x^4-4x^3+10x^2-12x+9\)
\(=x^2\left(x^2-2x+3\right)-2x\left(x^2-2x+3\right)+3\left(x^2-2x+3\right)\)
\(=\left(x^2-2x+3\right)^2\)
DH
0
MT
0
10 tháng 9 2021
\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\\ =-3\left(x^2-2\cdot\dfrac{5}{3}x+\dfrac{25}{9}-\dfrac{10}{9}\right)\\ =\dfrac{10}{3}-3\left(x-\dfrac{5}{3}\right)^2\\ =\left[\sqrt{\dfrac{10}{3}}-\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\left[\sqrt{\dfrac{10}{3}}+\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\\ =\left(\dfrac{\sqrt{30}}{3}+\dfrac{5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}}{3}-\dfrac{5\sqrt{3}}{3}+x\sqrt{3}\right)\)
\(=\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\)
10 tháng 9 2021
\(-3x^2+10x-5\)
\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{10}{9}\right)\)
\(=-3\left(x-\dfrac{5+\sqrt{10}}{3}\right)\left(x-\dfrac{5-\sqrt{10}}{3}\right)\)
