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\(2x^4-5x^3-27x^2+25x+50=0\)
\(\Leftrightarrow2x^4-4x^3-x^3+2x^2-25x^2+50x+25x^2-25x+50=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-x^2\left(x-2\right)-25x\left(x-5\right)+25\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-x^2-25x+25\right)=0\)
:D sorry mk ko bt phân tích 2x^3-x^2-25x+25 :D
b) 3x4-3x3+9x3-9x2-24x2+24x-48x+48
=3x3(x-1)+9x2(x-1)-24x(x-1)-48(x-1)
=(x-1)(3x3+9x2-24x-48)
=3(x-1)(x3+3x2-8x-16)
a) x4 - 10x3 - 15x2 + 20x + 4
= x4 + 2x3 - 12x3 - 24x2 + 9x2 + 18x + 2x + 4
= x3(x + 2) - 12x2(x + 2) + 9x(x + 2) + 2(x + 2)
= (x + 2)(x3 - 12x2 + 9x + 2)
b)
2x4 - 5x3 - 27x2 + 25x + 50
= 2x3(x - 2) - x2(x - 2) - 25x(x - 2) - 25(x - 2)
= (x - 2)(2x3 - x2 - 25x - 25)
c)\(3x^4+6x^3-33x^2-24x+48\)
\(=3\left(x^4+2x^3-11x^2-8x+16\right)\)
\(=3\left(x^4-x^3-4x^2+3x^3-3x^2-12x-4x^2+4x+16\right)\)
\(=3\left(x^2\left(x^2-x-4\right)+3x\left(x^2-x-4\right)-4\left(x^2-x-4\right)\right)\)
\(=3\left(x^2+3x-4\right)\left(x^2-x-4\right)\)
\(=3\left(x^2-x+4x-4\right)\left(x^2-x-4\right)\)
\(=3\left[x\left(x-1\right)+4\left(x-1\right)\right]\left(x^2-x-4\right)\)
\(=3\left(x-1\right)\left(x+4\right)\left(x^2-x-4\right)\)
a) 27x^3 –27x^2 +18x –4
= 27x^3 –9x^2–18x^2+6x + 12x –4
= 9x^2 (3x–1) – 6x (3x–1) +4(3x–1)
= (3x-1) (9x^2–6x+4)
b)2x^3–2x^2+5x+3
= 2x^3+x^2–2x^2–x+6+3
= x^2(2x+1)-x^2(2x+1)+3(2x+1)
= (2x+1) 3
c) 2x^4 + 5x^3+13x^2+25x+15
=2x^3(x+1)+3x^2(x+1)+10x(x+1)+15(x+1)
=(x+1)(x^2(2x+3)+5(2x+3))
=(x+1)(2x+3)(x^2+5)
a)2x2-6x=0
=>x(2x-6)=0
=>x=0 hoặc 2x-6=0
Với 2x-6=0 =>2x=6 <=>x=3