phân tích đa thức thành nhân tử
10x^2-28x-6
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10x(x-y)-(y-x)2
=10x(x-y)+(x-y).(y-x)
=(x-y).(10x+y-x)
=(x-y).(9x+y)
49x2+28x-5=49x2+28x+4-9=(7x+2)2-9=(7x+2-9)(7x+2+9)=(7x-7)(7x+11)=7(x-1)(7x+11)
\(x^4+8x^3+28x^2+48x-13\)
\(=x^4+4x^3+13x^2+4x^3+16x^2+52x-x^2-4x-13\)
\(=x^2\left(x^2+4x+13\right)+4x\left(x^2+4x+13\right)-\left(x^2+4x+13\right)\)
\(=\left(x^2+4x-1\right)\left(x^2+4x+13\right)\)
\(f\left(x\right)=x^4+8x^3+28x^2+48x-13\)
\(=\left(x^4+4x^3+7x^2\right)+\left(4x^3+16x^2+28x\right)+\left(5x^2+20x+35\right)-48\)
\(=x^2\left(x^2+4x+7\right)+4x\left(x^2+4x+7\right)+5\left(x^2+4x+7\right)-48\)
\(=\left(x^2+4x+7\right)\left(x^2+4x+5\right)-48\)
đặt t=\(x^2+4x+6\)khi đó g(t)=(t-1)(t+1)-48=t2-49=(t-7)(y+7)
vậy f(x)=(x2+4x-1)(x2+4x+13)
Trả lời:
Thay \(f\left(x\right)=0\), ta có:
\(0=x^4+8x^3+28x^2+48x-13\)
\(\Leftrightarrow-x^4-8x^3-28x^2-48x+13=0\)
\(\Leftrightarrow-x^4-4x^3-4x^3+x^2-16x^2-13x^2+4x-56x+13=0\)
\(\Leftrightarrow\left(-x^4-4x^3+x^2\right)+\left(-4x^3-16x^2+4x\right)+\left(-13x^2-56x+13\right)=0\)
\(\Leftrightarrow-x^2.\left(x^2+4x-1\right)-4x.\left(x^2+4x-1\right)-13.\left(x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(-x^2-4x-13\right).\left(x^2+4x-1\right)=0\)
Vì \(-x^2-4x-13=-x^2-4x-4-9\)
\(=-\left(x^2+4x+4\right)-9\)
\(=-\left(x+2\right)^2-9< 0\forall x\)
\(\Rightarrow x^2+4x-1=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)-5=0\)
\(\Leftrightarrow\left(x+2\right)^2=5=\left(\pm\sqrt{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=\sqrt{5}\\x+2=-\sqrt{5}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-2+\sqrt{5}\\x=-2-\sqrt{5}\end{cases}}\)
Vậy đa thức có 2 nghiêm \(x\in\left\{-2+\sqrt{5},-2-\sqrt{5}\right\}\)
a) \(5x^2y-20xy+20y=5y\left(x^2-4x+4\right)=5y\left(x-2\right)^2\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(3x^2y-12y=3y\left(x^2-4\right)=3y\left(x-2\right)\left(x+2\right)\)
d) \(7x^3-28x^2+28x=7x\left(x^2-4x+4\right)=7x\left(x-2\right)^2\)
a: \(5x^2y-20xy+20y\)
\(=4y\left(x^2-4x+4\right)\)
\(=4x\left(x-2\right)^2\)
b: \(3x^3+6x^2+3x\)
\(=3x\left(x^2+2x+1\right)\)
\(=3x\left(x+1\right)^2\)
c: \(3x^2y-12y\)
\(=3y\left(x^2-4\right)\)
\(=3y\left(x-2\right)\left(x+2\right)\)
d: \(7x^3-28x^2+28x\)
\(=7x\left(x^2-4x+4\right)\)
\(=7x\left(x-2\right)^2\)
a: \(12x^3-6x^2+3x\)
\(=3x\cdot4x^2-3x\cdot2x+3x\cdot1\)
\(=3x\left(4x^2-2x+1\right)\)
b: \(\dfrac{2}{5}x^2+5x^3+x^2y\)
\(=x^2\cdot\dfrac{2}{5}+x^2\cdot5x+x^2\cdot y\)
\(=x^2\left(\dfrac{2}{5}+5x+y\right)\)
c: \(14x^2y-21xy^2+28x^2y^2\)
\(=7xy\cdot2x-7xy\cdot3y+7xy\cdot4xy\)
\(=7xy\left(2x-3y+4xy\right)\)