brojevnids4

2022-12-14

What is the expansion of ${\left(x-1\right)}^{4}$?

Aryan Baxter

Expert

Determine the expansion of ${\left(x-1\right)}^{4}$.
We are aware that the binomial theorem says${\left(}^{a+b}=\sum k=0nckn·\left({a}^{n-k}{b}^{k}\right)$.
Figure out the expansion of ${\left(x-1\right)}^{4}$:
From the binomial theorem,
${\left(x-1\right)}^{4}=\sum k=04\frac{4!}{\left(4-k\right)!k!}{\left(x\right)}^{4-k}{\left(-1\right)}^{k}⇒{\left(x-1\right)}^{4}=\frac{4!}{\left(4-0\right)!0!}{\left(x\right)}^{4-0}{\left(-1\right)}^{0}+\frac{4!}{\left(4-1\right)!1!}{\left(x\right)}^{4-1}{\left(-1\right)}^{1}+\frac{4!}{\left(4-2\right)!2!}{\left(x\right)}^{4-2}{\left(-1\right)}^{2}+\frac{4!}{\left(4-3\right)!3!}{\left(x\right)}^{4-3}{\left(-1\right)}^{3}+\frac{4!}{\left(4-4\right)!4!}{\left(x\right)}^{4-4}{\left(-1\right)}^{4}⇒{\left(x-1\right)}^{4}=1{\left(x\right)}^{4}{\left(-1\right)}^{0}+4{\left(x\right)}^{3}{\left(-1\right)}^{1}+6{\left(x\right)}^{2}{\left(-1\right)}^{2}+4{\left(x\right)}^{1}{\left(-1\right)}^{3}+1{\left(x\right)}^{0}{\left(-1\right)}^{4}⇒{\left(x-1\right)}^{4}={x}^{4}-4{x}^{3}+6{x}^{2}-4x+1$
Hence, the required expansion is${\left(x-1\right)}^{4}={x}^{4}-4{x}^{3}+6{x}^{2}-4x+1$.

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