Giải bài 5 trang 28 SGK Toán 8 tập 1 - Cánh diều

Đề bài

Phân tích mỗi đa thức sau thành nhân tử:

a) \({\left( {x + 2y} \right)^2} - {\left( {x - y} \right)^2}\)                                            b) \({\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}\)

c) \(\left( {2y - 3} \right)x + 4y\left( {2y - 3} \right)\)                                      d) \(10{\rm{x}}\left( {x - y} \right) - 15{{\rm{x}}^2}\left( {y - x} \right)\)

e) \({x^3} + 3{{\rm{x}}^2} + 3{\rm{x}} + 1 - {y^3}\)                                         g) \({x^3} - 2{{\rm{x}}^2}y + x{y^2} - 4{\rm{x}}\)

Lời giải chi tiết

a) \(\begin{array}{l}{\left( {x + 2y} \right)^2} - {\left( {x - y} \right)^2}\\ = \left( {x + 2y + x - y} \right)\left( {x + 2y - x + y} \right)\\ = \left( {2{\rm{x}} + y} \right).3y\end{array}\)     

b)  \(\begin{array}{l}{\left( {x + 1} \right)^3} + {\left( {x - 1} \right)^3}\\ = \left( {x + 1 + x - 1} \right)\left[ {{{\left( {x + 1} \right)}^2} - \left( {x + 1} \right)\left( {x - 1} \right) + {{\left( {x - 1} \right)}^2}} \right]\\ = 2{\rm{x}}\left[ {{x^2} + 2{\rm{x}} + 1 - \left( {{x^2} - 1} \right) + {x^2} - 2{\rm{x}} + 1} \right]\\ = 2{\rm{x}}\left( {{x^2} + 2{\rm{x}} + 1 - {x^2} + 1 + {x^2} - 2{\rm{x}} + 1} \right)\\ = 2{\rm{x}}\left( {{x^2} + 3} \right)\end{array}\)        

c)  \(\left( {2y - 3} \right)x + 4y\left( {2y - 3} \right) = \left( {2y - 3} \right)\left( {1 + 4y} \right)\)   

d)      \(\begin{array}{l}10{\rm{x}}\left( {x - y} \right) - 15{{\rm{x}}^2}\left( {y - x} \right)\\ = 10{\rm{x}}\left( {x - y} \right) + 15{{\rm{x}}^2}\left( {x - y} \right)\\ = \left( {x - y} \right)\left( {10{\rm{x}} + 15{{\rm{x}}^2}} \right)\\ = \left( {x - y} \right).5{\rm{x}}\left( {2 + 3{\rm{x}}} \right)\\ = 5{\rm{x}}\left( {x - y} \right)\left( {2 + 3{\rm{x}}} \right)\end{array}\)   

e) \(\begin{array}{l}{x^3} + 3{{\rm{x}}^2} + 3{\rm{x}} + 1 - {y^3}\\ = \left( {{x^3} + 3{{\rm{x}}^2} + 3{\rm{x}} + 1} \right) - {y^3}\\ = {\left( {x + 1} \right)^3} - {y^3}\\ = \left( {x + 1 - y} \right)\left[ {{{\left( {x + 1} \right)}^2} + \left( {x + 1} \right)y + {y^2}} \right]\end{array}\)

g) \(\begin{array}{l}{x^3} - 2{{\rm{x}}^2}y + x{y^2} - 4{\rm{x}}\\{\rm{ = }}\left( {{x^3} - 2{{\rm{x}}^2}y + x{y^2}} \right) - 4{\rm{x}}\\ = x\left( {{x^2} - 2{\rm{x}}y + {y^2}} \right) - 4{\rm{x}}\\ = x{\left( {x - y} \right)^2} - 4{\rm{x}}\\ = x\left[ {{{\left( {x - y} \right)}^2} - {2^2}} \right]\\ = x\left( {x - y + 2} \right)\left( {x - y - 2} \right)\end{array}\)