Simplifying Radical Expressions worksheet with ten problems requiring simplification of radicals, including square roots and cube roots with variables and coefficients.

Let’s solve each radical expression one by one. We’ll simplify them by factoring out perfect powers (squares for square roots, cubes for cube roots) and using absolute value signs when needed — especially when taking even roots of variables that could be negative.

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Problem 1: ∛(448b⁷n²)

First, factor 448:
448 = 64 × 7 = 4³ × 7 → so ∛448 = ∛(64×7) = 4∛7

Now variables:
b⁷ = b⁶ × b = (b²)³ × b → ∛b⁷ = b²∛b
n² stays as is under cube root since it’s not a multiple of 3.

So:
∛(448b⁷n²) = ∛(64 × 7 × b⁶ × b × n²) = 4b² ∛(7bn²)

✔ Final: 4b²∛(7bn²)

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Problem 2: √(12p)

Factor 12: 12 = 4 × 3 → √12 = 2√3

So: √(12p) = √(4×3×p) = 2√(3p)

✔ Final: 2√(3p)

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Problem 3: √(36z²)

√36 = 6, √z² = |z| (because z could be negative, and square root gives non-negative result)

So: √(36z²) = 6|z|

✔ Final: 6|z|

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Problem 4: √8

√8 = √(4×2) = 2√2

✔ Final: 2√2

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Problem 5: ∛8

∛8 = 2, because 2³ = 8

✔ Final: 2

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Problem 6: ∛64

64 = 4³ → ∛64 = 4

✔ Final: 4

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Problem 7: ∛(64z³)

∛64 = 4, ∛z³ = z (no absolute value needed for odd roots)

So: 4z

✔ Final: 4z

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Problem 8: √112

Factor 112: 112 = 16 × 7 → √112 = √(16×7) = 4√7

✔ Final: 4√7

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Problem 9: √(275b⁵h⁶)

Factor 275: 275 = 25 × 11 → √275 = 5√11

Variables:
b⁵ = b⁴ × b → √b⁵ = b²√b
h⁶ = (h³)² → √h⁶ = h³

So: √(275b⁵h⁶) = 5b²h³√(11b)

✔ Final: 5b²h³√(11b)

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Problem 10: ∛(135y⁷p²)

Factor 135: 135 = 27 × 5 = 3³ × 5 → ∛135 = 3∛5

Variables:
y⁷ = y⁶ × y = (y²)³ × y → ∛y⁷ = y²∛y
p² stays under cube root

So: ∛(135y⁷p²) = 3y² ∛(5yp²)

✔ Final: 3y²∛(5yp²)

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Final Answer:
1) 4b²∛(7bn²)
2) 2√(3p)
3) 6|z|
4) 2√2
5) 2
6) 4
7) 4z
8) 4√7
9) 5b²h³√(11b)
10) 3y²∛(5yp²)