Simplify. Assume s is greater than or equal to zero. sqrt 245s^8 ______

Factorize 245 and s^8: sqrt(245s^8) First, let's factor 245 into its prime factors and express s^8 in terms of s squared to the fourth power. Prime factorization of 245 is 5 * 7 * 7. s^8 can be written as (s^2)^4. sqrt(245s^8) = sqrt(5 * 7 * 7 * (s^2)^4)Group Identical Factors: sqrt(5 * 7 * 7 * (s^2)^4) Now, group the identical factors and use the exponents. sqrt(5 * 7^2 * (s^2)^4)Apply Product Property: sqrt(5 * 7^2 * (s^2)^4) Product property of radicals: sqrt(a * b) = sqrt(a) * sqrt(b) sqrt(5 * 7^2 * (s^2)^4) = sqrt(5) * sqrt(7^2) * sqrt((s^2)^4)Cancel Square Roots: sqrt(5) * sqrt(7^2) * sqrt((s^2)^4) Square and square root cancel each other out for sqrt(7^2) and sqrt((s^2)^4). sqrt(5) * 7 * (s^2)^2 = sqrt(5) * 7 * s^4 = 7s^4sqrt(5)

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Simplify. Assume $s$ is greater than or equal to zero. $\newline$ $\sqrt{245s^8}$

View step-by-step help

Home

Math Problems

Algebra 1

Simplify radical expressions: mixed review

Full solution

Q. Simplify. Assume

s

is greater than or equal to zero.

\newline

\sqrt{245s^8}

Factorize $245$ and $s^8$ : $\sqrt{245s^8}$ $\newline$ First, let's factor $245$ into its prime factors and express $s^8$ in terms of $s$ squared to the fourth power. $\newline$ Prime factorization of $245$ is $5 \times 7 \times 7$ . $\newline$ $s^8$ can be written as $(s^2)^4$ . $\newline$ $\sqrt{245s^8} = \sqrt{5 \times 7 \times 7 \times (s^2)^4}$
Group Identical Factors: $\sqrt{5 \times 7 \times 7 \times (s^2)^4}$ $\newline$ Now, group the identical factors and use the exponents. $\newline$ $\sqrt{5 \times 7^2 \times (s^2)^4}$
Apply Product Property: $\sqrt{5 \times 7^2 \times (s^2)^4}$ $\newline$ Product property of radicals: $\newline$ $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ $\newline$ $\sqrt{5 \times 7^2 \times (s^2)^4} = \sqrt{5} \times \sqrt{7^2} \times \sqrt{(s^2)^4}$
Cancel Square Roots: $\sqrt{5} \times \sqrt{7^2} \times \sqrt{(s^2)^4}$ $\newline$ Square and square root cancel each other out for $\sqrt{7^2}$ and $\sqrt{(s^2)^4}$ . $\newline$ $\sqrt{5} \times 7 \times (s^2)^2$ $\newline$ = $\sqrt{5} \times 7 \times s^4$ $\newline$ = $7s^4\sqrt{5}$