Toán Lớp 8: phân tích đa thức thành nhân tử
b, y.(x^3+8)-y.(x+2).(x+2).(5-2x)
c, 5x^4-45x^2y^2-20x^3+20x^2
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Toán Lớp 8: phân tích đa thức thành nhân tử
b, y.(x^3+8)-y.(x+2).(x+2).(5-2x)
c, 5x^4-45x^2y^2-20x^3+20x^2
Comments ( 1 )
b)y\left( {{x^3} + 8} \right) – y\left( {x + 2} \right)\left( {x + 2} \right)\left( {5 – 2x} \right)\\
= y\left( {x + 2} \right)\left( {{x^2} – 2x + 4} \right) – y\left( {x + 2} \right)\left( {5x – 2{x^2} + 10 – 4x} \right)\\
= y\left( {x + 2} \right)\left( {{x^2} – 2x + 4 – 5x + 2{x^2} – 10 + 4x} \right)\\
= y\left( {x + 2} \right).\left( {3{x^2} – 3x – 6} \right)\\
= 3y\left( {x + 2} \right)\left( {{x^2} – x – 2} \right)\\
= 3y\left( {x + 2} \right)\left( {x – 2} \right)\left( {x + 1} \right)\\
c)5{x^4} – 45{x^2}{y^2} – 20{x^3} + 20{x^2}\\
= 5{x^2}\left( {{x^2} – 9{y^2} – 4x + 4} \right)\\
= 5{x^2}\left[ {{{\left( {x – 2} \right)}^2} – 9{y^2}} \right]\\
= 5{x^2}\left( {x – 2 – 3y} \right)\left( {x – 2 + 3y} \right)
\end{array}$