Phân tích đa thức thành nhân tử

N

nguyenbahiep1

a)

[tex](x^2+4x+8)^2+3x^3+12x^2+24x \\ (x^2+4x+8)^2+3x(x^2+4x^2+8) = (x^2+4x+8)(x^2+4x+8 +3x) = (x^2+4x+8)(x^2+7x+8) [/tex]



b)

[tex](x^2+1)^4+9(x^2+1)^3+21(x^2+1)-x^2-31 \\ (x^2+1)^4+9(x^2+1)^3+21(x^2+1)-(x^2+1)-30 \\ (x^2+1)^4 - (x^2+1)^3 + 10(x^2+1)^3 - 10(x^2+1) + 30(x^2+1) -30 \\ (x^2+1)^3(x^2+1-1) + 10(x^2+1)( (x^2+1)^2-1^2) + 30(x^2+1-1) \\ (x^2+1)^3(x^2) + 10.x^2.(x^2+2) + 30x^2 = x^2( (x^2+1)^3 + 10(x^2+2) + 30 ) [/tex]



c)

[TEX](x+1)(x+2)(x+4)(x+5)-40 \\ (x^2+3x+2)(x^2+9x+20) -40 \\ x^4 + 12x^3+49x^2+18x \\ x( x^3+12x^2+49x+18)[/TEX]
 
Last edited by a moderator:
H

harrypham

c) [TEX](x+1)(x+2)(x+4)(x+5)-40[/TEX]
[TEX]=[(x+1)(x+5)] \times [(x+2)(x+4)]-40[/TEX]
[TEX]=(x^2+6x+5)(x^2+6x+8)-40[/TEX]
[TEX]= \left( x^2+6x+ \frac{13}{2}- \frac{3}{2} \right) \left(x^2+6x+ \frac{13}{2}+ \frac{3}{2} \right)-40[/TEX]
[TEX]= \left( x^2+6x+ \frac{13}{2} \right)- \frac{9}{4}-40[/TEX]
[TEX]= \left( x^2+6x+ \frac{13}{2} \right) - \frac{169}{4}[/TEX]
[TEX]= \left( x^2+6x+\frac{13}{2}+ \frac{13}{2} \right) \left( x^2+6x+ \frac{13}{2}- \frac{13}{2} \right)[/TEX]
[TEX]= (x^2+6x+13)x(x+6)[/TEX].
 
Last edited by a moderator:
Top Bottom