Respuesta :

kaevras
I used bionomial expansion up to and including the term [tex] x^{3} [/tex]

[tex] (-3 + x)^{5} [/tex]

[tex] -3^{5} (1 + -\frac{1}{3}x) ^{5} [/tex]

[tex] -243 (1 + -\frac{1}{3}x) ^{5} [/tex]

n = 5
x = [tex]- \frac{1}{3}x [/tex]


Binomial expansion:

[tex]1 + nx + \frac{n(n-1) x^{2} }{2!} + \frac{n(n-1)(n-2) x^{3} }{3!} [/tex]


Therefore:

[tex]1 + (5*- \frac{1}{3}x) + \frac{5(4) (- \frac{1}{3}x)^{2} }{2} + \frac{5(4)(3) (- \frac{1}{3}x)^{3} }{6} [/tex]

=

[tex]1 - \frac{5}{3} x + \frac{10}{9} x^{2} - \frac{10}{27} x^{3} [/tex]

Multiply by -243

=

[tex]-243 + 405x - 270x^{2} + 90 x^{3} [/tex]

Answer:

[tex]x^5-15x^4+90x^3-270x^2+405x-243[/tex]

Step-by-step explanation:

Here, the given expression is,

[tex](x-3)^5[/tex]

We know that by binomial theorem,

[tex](p+q)^n=\sum_{r=0}^{n} ^nC_r (p)^{n-r} (q)^r[/tex]

Where,

[tex]^nC_r=\frac{n!}{r!(n-r)!}[/tex]

Now,

[tex](x-3)^5=(x+(-3))^5[/tex]

Thus, by the above theorem,

[tex](x+(-3))^5=\sum_{r=0}^{5} ^5C_r (x)^{5-r} (-3)^r[/tex]

[tex]=^5C_0 (x)^{5-0} (-3)^0+^5C_1 (x)^{5-1} (-3)^1+^5C_2 (x)^{5-2} (-3)^2+^5C_3 (x)^{5-3} (-3)^3+^5C_4 (x)^{5-4} (-3)^4+^5C_5 (x)^{5-5} (-3)^5[/tex]

[tex]=(x)^5(1)+5(x)^4(-3)+10(x)^3(-3)^2+10(x)^2(-3)^3+5(x)(-3)^4+(-3)^5[/tex]

[tex]=x^5-15x^4+90x^3-270x^2+405x-243[/tex]

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