Bài 13 trang 34 Tài liệu dạy – học Toán 8 tập 1

Đề bài

Phân tích các đa thức sau thành nhân tử

a) \({x^2} - 3x + 2\) ;

b) \({x^2} - x - 6\) ;

c) \({x^2} + 7x + 12\) ;

d) \(3{x^2} + x - 2\) ;

e) \({x^4} + {x^2} + 1\) ;

f) \({x^2} + 2xy - 15{y^2}\) .

Lời giải chi tiết

\(\eqalign{ & a)\,\,{x^2} - 3x + 2 \cr & \,\,\,\,\, = {x^2} - x - 2x + 2 \cr & \,\,\,\, = \left( {{x^2} - x} \right) - \left( {2x - 2} \right) \cr & \,\,\,\, = x\left( {x - 1} \right) - 2\left( {x - 1} \right) \cr & \,\,\,\, = \left( {x - 1} \right)\left( {x - 2} \right) \cr & b)\,\,{x^2} - x - 6 \cr & \,\,\,\,\, = {x^2} - 3x + 2x - 6 \cr & \,\,\,\,\, = \left( {{x^2} - 3x} \right) + \left( {2x - 6} \right) \cr & \,\,\,\,\, = x\left( {x - 3} \right) + 2\left( {x - 3} \right) \cr & \,\,\,\,\, = \left( {x - 3} \right)\left( {x + 2} \right) \cr & c)\,\,{x^2} + 7x + 12 \cr & \,\,\,\,\, = {x^2} + 3x + 4x + 12 \cr & \,\,\,\,\, = \left( {{x^2} + 3x} \right) + \left( {4x + 12} \right) \cr & \,\,\,\,\, = x\left( {x + 3} \right) + 4\left( {x + 3} \right) \cr & \,\,\,\,\, = \left( {x + 3} \right)\left( {x + 4} \right) \cr & d)\,\,3{x^2} + x - 2 \cr & \,\,\,\,\, = 3{x^2} + 3x - 2x - 2 \cr & \,\,\,\,\, = \left( {3{x^2} + 3x} \right) - \left( {2x + 2} \right) \cr & \,\,\,\,\, = 3x\left( {x + 1} \right) - 2\left( {x + 1} \right) \cr & \,\,\,\,\, = \left( {x + 1} \right)\left( {3x - 2} \right) \cr & e)\,\,{x^4} + {x^2} + 1 \cr & \,\,\,\,\, = {x^4} + {x^2} + 1 + {x^2} - {x^2} \cr & \,\,\,\,\, = \left( {{x^4} + 2{x^2} + 1} \right) - {x^2} \cr & \,\,\,\,\, = {\left( {{x^2} + 1} \right)^2} - {x^2} \cr & \,\,\,\,\, = \left( {{x^2} + 1 - x} \right)\left( {{x^2} + 1 + x} \right) \cr & f)\,\,{x^2} + 2xy - 15{y^2} \cr & \,\,\,\,\, = {x^2} + 2xy + {y^2} - 16{y^2} \cr & \,\,\,\,\, = \left( {{x^2} + 2xy + {y^2}} \right) - 16{y^2} \cr & \,\,\,\,\, = {\left( {x + y} \right)^2} - {\left( {4y} \right)^2} \cr & \,\,\,\,\, = \left( {x + y - 4y} \right)\left( {x + y + 4y} \right) \cr & \,\,\,\,\, = \left( {x - 3y} \right)\left( {x + 5y} \right) \cr} \)

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