Bài 57 trang 14 SBT toán 8 tập 1

Đề bài

Phân tích các đa thức sau thành nhân tử:

\(a)\) \({x^3} - 3{x^2} - 4x + 12\)

\(b)\) \({x^4} - 5{x^2} + 4\)

\(c)\) \({\left( {x + y + z} \right)^3} - {x^3} - {y^3} - {z^3}\)

Lời giải

\(a)\) \({x^3} - 3{x^2} - 4x + 12\) \( = \left( {{x^3} - 3{x^2}} \right) - \left( {4x - 12} \right)\)

\( = {x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right)\)

\( = \left( {x - 3} \right)\left( {{x^2} - 4} \right) \)

\(= \left( {x - 3} \right)\left( {x + 2} \right)\left( {x - 2} \right)\)

\(b)\) \({x^4} - 5{x^2} + 4\)

\( = {x^4} - 4{x^2} - {x^2} + 4 \)

\(= \left( {{x^4} - 4{x^2}} \right) - \left( {{x^2} - 4} \right)\)

\( = {x^2}\left( {{x^2} - 4} \right) - \left( {{x^2} - 4} \right) \)

\(= \left( {{x^2} - 4} \right)\left( {{x^2} - 1} \right)\)

\( = \left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 1} \right)\left( {x - 1} \right)\)

\(c)\) \({\left( {x + y + z} \right)^3} - {x^3} - {y^3} - {z^3}\)

\( = {\left[ {\left( {x + y} \right) + z} \right]^3} - {x^3} - {y^3} - {z^3}\)

\( = {\left( {x + y} \right)^3} + 3{\left( {x + y} \right)^2}z\)\( + 3\left( {x + y} \right){z^2} + {z^3} - {x^3} - {y^3} - {z^3}\)

\(= {x^3} + {y^3} + 3xy\left( {x + y} \right) + 3{\left( {x + y} \right)^2}z\)\(+ 3\left( {x + y} \right){z^2} - {x^3} - {y^3} \)

\(= 3\left( {x + y} \right)\left[ {xy + \left( {x + y} \right)z + {z^2}} \right] \)

\(= 3\left( {x + y} \right)\left[ {xy + xz + yz + {z^2}} \right] \)

\( = 3\left( {x + y} \right)\left[ {x\left( {y + z} \right) + z\left( {y + z} \right)} \right]\)

\( = 3\left( {x + y} \right)\left( {y + z} \right)\left( {x + z} \right) \)


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