Question

Expand the following using either the Binomial Theorem or Pascal’s Triangle. You must show your work for credit. (x - 3)^5

Answer 1

I used bionomial expansion up to and including the term $x^{3}$

$(-3 + x)^{5}$

$-3^{5} (1 + -\frac{1}{3}x) ^{5}$

$-243 (1 + -\frac{1}{3}x) ^{5}$

n = 5
x = $- \frac{1}{3}x$


Binomial expansion:

$1 + nx + \frac{n(n-1) x^{2} }{2!} + \frac{n(n-1)(n-2) x^{3} }{3!}$


Therefore:

$1 + (5*- \frac{1}{3}x) + \frac{5(4) (- \frac{1}{3}x)^{2} }{2} + \frac{5(4)(3) (- \frac{1}{3}x)^{3} }{6}$

=

$1 - \frac{5}{3} x + \frac{10}{9} x^{2} - \frac{10}{27} x^{3}$

Multiply by -243

=

$-243 + 405x - 270x^{2} + 90 x^{3}$

Answer 2

Answer:

$x^5-15x^4+90x^3-270x^2+405x-243$

Step-by-step explanation:

Here, the given expression is,

$(x-3)^5$

We know that by binomial theorem,

$(p+q)^n=\sum_{r=0}^{n} ^nC_r (p)^{n-r} (q)^r$

Where,

$^nC_r=\frac{n!}{r!(n-r)!}$

Now,

$(x-3)^5=(x+(-3))^5$

Thus, by the above theorem,

$(x+(-3))^5=\sum_{r=0}^{5} ^5C_r (x)^{5-r} (-3)^r$

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

$=(x)^5(1)+5(x)^4(-3)+10(x)^3(-3)^2+10(x)^2(-3)^3+5(x)(-3)^4+(-3)^5$

$=x^5-15x^4+90x^3-270x^2+405x-243$