SOLUTION
The given expression is:
[tex](x-4)^5[/tex]
Using binomial theorem, the function is expanded as follows:
[tex](x-4)^5=x^5+5(x)^4(-4)^+\frac{5(5-1)}{2!}x^3(-4)^2+\frac{5(5-1)(5-2)}{3!}x^2(-4)^3+\frac{5(5-1)(5-2)(5-3)}{4!}x^(-4)^4+(-4)^5[/tex]
This gives:
[tex](x-4)^5=x^5-20x^4+160x^3-640x^2+1280x-1024[/tex]
The pascal triangle is shown:
Using pascal triangle the expansion is shown:
[tex]\begin{gathered} (x-4)^5=x^5+5x^4(-4)+10x^3(-4)^2+10x^2(-4)^3+5x(-4)^4+(-4)^5 \\ (x-4)^5=x^5-20x^4+160x^3-640x^2+1280x-1024 \end{gathered}[/tex]
