Answer:
[tex]x^5-25x^4+250x^3-1250x^2+3125 x-3125[/tex]
Step-by-step explanation:
Here the given expression is,
[tex](x-5)^5[/tex]
By the binomial theorem,
[tex](p+q)^n=\sum_{r=0}^{n} ^nC_rp^{n-r} q^r[/tex]
Where,
[tex]^nC_r=\frac{n!}{(n-r)!r!}[/tex]
Thus, by the above formula,
[tex](x-5)^5=\sum_{r=0}^{5} ^5C_rx^{5-r} (-5)^r[/tex]
[tex]=^5C_0x^{5-0} (-5)^0 + ^5C_1x^{5-1} (-5)^1 + ^5C_2x^{5-2} (-5)^2 + ^5C_3x^{5-3} (-5)^3 + ^5C_4x^{5-4} (-5)^4 + ^5C_5x^{5-5} (-5)^5[/tex]
[tex]=(1)(x^5)(1)+(5)(x^4)(-5)+(10)(x^3)(25)+(10)(x^2)(-125)+(5)(x^1)(625)+(1)(1)(-3125)[/tex]
[tex]=x^5-25x^4+250x^3-1250x^2+3125 x-3125[/tex]
