# How do you use the binomial theorem to expand (sqrtx+3)^4?

Jan 6, 2018

$= {x}^{2} + 12 x \sqrt{x} + 54 x + 108 \sqrt{x} + 81$

#### Explanation:

row in pascal's triangle for powers of $4$:

1 4 6 4 1

for ${\left(x + y\right)}^{4}$, this means $1 \left({x}^{4} {y}^{0}\right) + 4 \left({x}^{3} {y}^{1}\right) + 6 \left({x}^{2} {y}^{2}\right) + 4 \left({x}^{1} {y}^{3}\right) + 1 \left({x}^{0} {y}^{4}\right)$,

or ${x}^{4} + 4 {x}^{3} y + 6 {x}^{2} {y}^{2} + 4 x {y}^{3} + {y}^{4}$

for ${\left(\sqrt{x} + 3\right)}^{4} :$

${\left(\sqrt{x}\right)}^{4} + 4 \left({\left(\sqrt{x}\right)}^{3} \cdot 3\right) + 6 \left(x \cdot 9\right) + 4 \left(\left(\sqrt{x}\right) \cdot 27\right) + 81$

$= {x}^{2} + 12 x \sqrt{x} + 54 x + 108 \sqrt{x} + 81$

or ${x}^{2} + 12 {x}^{\frac{3}{2}} + 54 x + 108 {x}^{\frac{1}{2}} + 81$