Question

How do you expand (x-3)^5 using the binomial theorem?

Answer 1

Binomial theorem:
\[(a+b)^n=\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}a^{n-k}b^k\]

Answer 2

Where \(\begin{pmatrix}n\\k\end{pmatrix}=\dfrac{n!}{k!(n-k)!}\).

Answer 3

i dont understand that equation

Answer 4

So for this problem, you have
\[\begin{align*}(x-3)^5&=\sum_{k=0}^5\begin{pmatrix}5\\k\end{pmatrix}x^{5-k}(-3)^k\\
&=\begin{pmatrix}5\\0\end{pmatrix}x^{5-0}(-3)^0+\begin{pmatrix}5\\1\end{pmatrix}x^{5-1}(-3)^1+\begin{pmatrix}5\\2\end{pmatrix}x^{5-2}(-3)^2\\
&~~~~+\begin{pmatrix}5\\3\end{pmatrix}x^{5-3}(-3)^3+\begin{pmatrix}5\\4\end{pmatrix}x^{5-4}(-3)^4+\begin{pmatrix}5\\5\end{pmatrix}x^{5-5}(-3)^5
\\&=\cdots\end{align*}\]

Answer 5

so would my final answer be x^5+x^4+x^3+x^2+x-183?

Answer 6

No. You're forgetting the coefficients. Are you familiar with the binomial coefficient?

Answer 7

no
I dont know what that is?

Answer 8

is it 1, 5 10 10 5 1?

Answer 9

Yes! Also, you have to keep track of the \((-3)^{\cdots}\). They also contribute to the coefficients of each power of x.

Answer 10

For example, I'll work out the third term:
\[\begin{pmatrix}5\\2\end{pmatrix}x^{5-2}(-3)^2=5x^3(-9)=-45x^3\]

Answer 11

ok is it x^5+5x^4+10x^3+10x^2+5x+24267?

Answer 12

No, you're still missing the powers of (-3) in each term.

Answer 13

Would the first term is x^5 (-3^0) right? and the second would be x^4 (-3^1)*5?
and so on

Answer 14

So would the final answer be x^5-15x^4+90x^3-270x^2+405x-243?

Answer 15

Yes!

Answer 16

YESS Thankyou very much

Answer 17

You're welcome