The binomial theorem tells us how to expand an expression of the form (a + b)^2 like this:
[tex](a+b)^n=nC_0a^nb^0+nC_1a^{n-1}b^1+\cdots nC_na^0b^n[/tex]And nCr is given by the following formula:
[tex]nC_r=\frac{n!}{r!(n-r)!}[/tex]Then, for this polynomial, we can apply the binomial theorem with a = x, b = -3 and n = 5 to get:
[tex](x-3)^5=5C_0x^5(-3)^0+5C_1x^4(-3)^1+5C_2x^3(-3)^2+5C_3x^2(-3)^3+5C_4x^1(-3)^4+5C_5x^0(-3)^5[/tex][tex]\begin{gathered} 5C_0=1 \\ 5C_1=5 \\ 5C_2=10 \\ 5C_3=10 \\ 5C_4=5 \\ 5C_5=1 \end{gathered}[/tex]Simplifying, we get:
[tex]\begin{gathered} (x-3)^5=(1)x^5(-3)^0+(5)_{}x^4(-3)^1+(10)_{}x^3(-3)^2+(10)x^2(-3)^3+(5)x^1(-3)^4+(1)x^0(-3)^5 \\ (x-3)^5=x^5+(5)_{}x^4(-3)^{}+(10)_{}x^3(9)+(10)x^2(-27)+(5)x^1(81)^{}-243 \\ (x-3)^5=x^5-15_{}x^4^{}+90_{}x^3-270x^2+405x^{}^{}-243 \end{gathered}[/tex]Then, the expanded polynomial is:
x⁵ - 15x⁴ + 90x³ - 270x² + 405x - 243
