Find p(q(x)) p(x)=x^2+3 q(x) = square root of x+2

$p (x) = x^{2} + 3$ $q (x) = \sqrt{x + 2}$

Set up the composite result function.

$p (q (x))$

Evaluate $p (\sqrt{x + 2})$ by substituting in the value of $q$ into $p$ .

$p (\sqrt{x + 2}) = {(\sqrt{x + 2})}^{2} + 3$

Rewrite ${\sqrt{x + 2}}^{2}$ as $x + 2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 2}$ as ${(x + 2)}^{\frac{1}{2}}$ .

$p (\sqrt{x + 2}) = {({(x + 2)}^{\frac{1}{2}})}^{2} + 3$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$p (\sqrt{x + 2}) = {(x + 2)}^{\frac{1}{2} \cdot 2} + 3$

Combine $\frac{1}{2}$ and $2$ .

$p (\sqrt{x + 2}) = {(x + 2)}^{\frac{2}{2}} + 3$

Cancel the common factor of $2$ .

Cancel the common factor.

$p (\sqrt{x + 2}) = {(x + 2)}^{\frac{2}{2}} + 3$

Rewrite the expression.

$p (\sqrt{x + 2}) = (x + 2) + 3$

Simplify.

$p (\sqrt{x + 2}) = x + 2 + 3$

Add $2$ and $3$ .

$p (\sqrt{x + 2}) = x + 5$