# How do you use the Binomial Theorem to expand (x + 1)^4?

May 2, 2018

${x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 1$

#### Explanation:

The binomial theorem states:
${\left(a + b\right)}^{4} = {a}^{4} + 4 {a}^{3} b + 6 {a}^{2} {b}^{2} + 4 a {b}^{3} + {b}^{4}$

so here, $a = x \mathmr{and} b = 1$

We get:
${\left(x + 1\right)}^{4} = {x}^{4} + 4 {x}^{3} \left(1\right) + 6 {x}^{2} {\left(1\right)}^{2} + 4 x {\left(1\right)}^{3} + {\left(1\right)}^{4}$
${\left(x + 1\right)}^{4} = {x}^{4} + 4 {x}^{3} + 6 {x}^{2} + 4 x + 1$

May 2, 2018

$1 + 4 x + 6 {x}^{2} + 4 {x}^{3} + {x}^{4}$

#### Explanation:

Binomial expansion is given by:
(a+bx)^n=sum_(r=0)^n(n!)/(r!(n-r)!)a^(n-r)(bx)^r

So, for ${\left(1 + x\right)}^{4}$ we have:
(4!)/(0!(4-0)!)1^(4-0)x^0+(4!)/(1!(4-1)!)1^(4-1)x^1+(4!)/(2!(4-2)!)1^(4-2)x^2+(4!)/(3!(4-3)!)1^(4-3)x^3+(4!)/(4!(4-4)!)1^(4-4)x^4

$1 + 4 x + 6 {x}^{2} + 4 {x}^{3} + {x}^{4}$