Bài 57 trang 25 SGK Toán 8 tập 1

Phân tích các đa thức sau thành nhân tử:

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

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

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

d) \({x^4} + 4\)

(Gợi ý câu d): Thêm và bớt \(4{x^2}\) vào đa thức đã cho).

a) 

\(\eqalign{
& {x^2}-4x + 3 = {x^2}-x - 3x + 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\left( {{x^2} - x} \right) + \left( { - 3x + 3} \right)\cr&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= x\left( {x - 1} \right) - 3\left( {x - 1} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {x - 1} \right)\left( {x - 3} \right) \cr} \)

b) 

\(\eqalign{
& {x^2} + 5x + 4 = {x^2} + 4x + x + 4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \, = \left( {{x^2} + 4x} \right) + \left( {x + 4} \right)\cr&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \, = x\left( {x + 4} \right) + \left( {x + 4} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {x + 4} \right)\left( {x + 1} \right) \cr} \)

c) 

\(\eqalign{
& {x^2}-x-6 = {x^2} + 2x-3x-6 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\,= \left( {{x^2} + 2x} \right) + \left( { - 3x - 6} \right)\cr&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\, = x\left( {x{\rm{ }} + {\rm{ }}2} \right) - 3\left( {x + 2} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\, = \left( {x + 2} \right)\left( {x - 3} \right) \cr} \)

d)

\(\eqalign{
& {x^4} + 4 = {x^4} + 4{x^2} + 4-4{x^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;  = \left( {{x^4} + 4{x^2} + 4} \right) - 4{x^2}\cr&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; = \left[ {{{\left( {{x^2}} \right)}^2} + 2.{x^2}.2 + {2^2}} \right] - 4{x^2}\,\,\,\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; = {({x^2} + 2)^2}-{\left( {2x} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\; = ({x^2} + 2-2x)({x^2} + 2 + 2x) \cr} \)