Find the Degree f(x) = square root of x( square root of x-6)

$f (x) = \sqrt{x} (\sqrt{x} - 6)$

Step 1

Simplify and reorder the polynomial.

Apply the distributive property.

$\sqrt{x} \sqrt{x} + \sqrt{x} \cdot - 6$

Multiply $\sqrt{x} \sqrt{x}$ .

Raise $\sqrt{x}$ to the power of $1$ .

${\sqrt{x}}^{1} \sqrt{x} + \sqrt{x} \cdot - 6$

Raise $\sqrt{x}$ to the power of $1$ .

${\sqrt{x}}^{1} {\sqrt{x}}^{1} + \sqrt{x} \cdot - 6$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

${\sqrt{x}}^{1 + 1} + \sqrt{x} \cdot - 6$

Add $1$ and $1$ .

${\sqrt{x}}^{2} + \sqrt{x} \cdot - 6$

Move $- 6$ to the left of $\sqrt{x}$ .

${\sqrt{x}}^{2} - 6 \cdot \sqrt{x}$

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

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

${(x^{\frac{1}{2}})}^{2} - 6 \cdot \sqrt{x}$

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

$x^{\frac{1}{2} \cdot 2} - 6 \cdot \sqrt{x}$

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

$x^{\frac{2}{2}} - 6 \cdot \sqrt{x}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{\frac{2}{2}} - 6 \cdot \sqrt{x}$

Rewrite the expression.

$x^{1} - 6 \cdot \sqrt{x}$

Simplify.

$x - 6 \cdot \sqrt{x}$

$x - 6 \sqrt{x}$

Step 2

The degree cannot be determined because $x - 6 \sqrt{x}$ is not a polynomial.

Not a polynomial