# How do you simplify sqrt(63c^3d^4f^5)?

Mar 21, 2017

$3 c {d}^{2} {f}^{2} \cdot \sqrt{7 c f}$

#### Explanation:

$\sqrt{63 {c}^{3} {d}^{4} {f}^{5}}$

$\therefore = {\left(63 {c}^{3} {d}^{4} {f}^{5}\right)}^{\frac{1}{2}}$

$\therefore = {\left(7 \cdot 3 \cdot 3 \cdot {c}^{3} {d}^{4} {f}^{5}\right)}^{\frac{1}{2}}$

$\therefore = {\left(7 \cdot {3}^{2} {c}^{3} {d}^{4} {f}^{5}\right)}^{\frac{1}{2}}$

$\therefore = \left({7}^{\frac{1}{2}} \cdot {3}^{2 \times \frac{1}{2}} {c}^{3 \times \frac{1}{2}} {d}^{4 \times \frac{1}{2}} {f}^{5 \times \frac{1}{2}}\right)$

$\therefore = \left({7}^{\frac{1}{2}} \cdot {3}^{1} \cdot {c}^{\frac{3}{2}} \cdot {d}^{\frac{4}{2}} \cdot {f}^{\frac{5}{2}}\right)$

$\therefore = \sqrt{7} \cdot 3 \cdot \sqrt{{c}^{3}} \cdot \sqrt{{d}^{4}} \cdot \sqrt{{f}^{5}}$

:.=3d^2*sqrt(7c^3f^5

$\therefore = 3 {d}^{2} \sqrt{7 \cdot c \cdot c \cdot c \cdot f \cdot f \cdot f \cdot f \cdot f}$

$\sqrt{a} \cdot \sqrt{a} = a$

$\therefore = 3 c {d}^{2} {f}^{2} \sqrt{7 c f}$

Mar 21, 2017

$3 c {d}^{2} {f}^{2} \sqrt{7 c f}$

#### Explanation:

Given$: \text{ } \sqrt{63 {c}^{3} {d}^{4} {f}^{5}}$

As soon as you see a square root you know that you are looking for squared values within that root. You can 'take outside' the root all those squared values.

Write as:$\text{ } \sqrt{{3}^{2} \times 7 \times {c}^{2} \times c \times {\left({d}^{2}\right)}^{2} \times {\left({f}^{2}\right)}^{2} \times f}$

Taking the squared values 'outside' the root giving:

$3 c {d}^{2} {f}^{2} \sqrt{7 c f}$