Multiplying-Radical-Expressions Worksheet | PDF - Free Printable

Let's solve each problem step-by-step from the worksheet titled "Multiplying Radical Expressions".

We'll use the following key rules for radicals:

1. $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$
2. $\sqrt{a^2} = a$ (for $a \geq 0$)
3. Distributive property: $a(b + c) = ab + ac$
4. $(a + b)(c + d) = ac + ad + bc + bd$
5. Simplify square roots by factoring perfect squares.

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Problems and Solutions

#### 1) $\sqrt{5} \times \sqrt{5}$
$$
= \sqrt{5 \cdot 5} = \sqrt{25} = 5
$$

#### 2) $\sqrt{5} \times \sqrt{10}$
$$
= \sqrt{5 \cdot 10} = \sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}
$$

#### 3) $\sqrt{2} \times \sqrt{18}$
$$
= \sqrt{2 \cdot 18} = \sqrt{36} = 6
$$

#### 4) $\sqrt{14} \times \sqrt{21}$
$$
= \sqrt{14 \cdot 21} = \sqrt{294}
$$
Factor: $294 = 49 \cdot 6 = 7^2 \cdot 6$, so:
$$
= \sqrt{49 \cdot 6} = 7\sqrt{6}
$$

#### 5) $\sqrt{5} \times -4\sqrt{20}$
First, simplify $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}$
So:
$$
= \sqrt{5} \cdot (-4 \cdot 2\sqrt{5}) = \sqrt{5} \cdot (-8\sqrt{5}) = -8 \cdot (\sqrt{5} \cdot \sqrt{5}) = -8 \cdot 5 = -40
$$

#### 6) $3\sqrt{12} \times \sqrt{6}$
Simplify $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$
So:
$$
= 3 \cdot 2\sqrt{3} \cdot \sqrt{6} = 6\sqrt{3} \cdot \sqrt{6} = 6\sqrt{18}
$$
Now $\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$, so:
$$
= 6 \cdot 3\sqrt{2} = 18\sqrt{2}
$$

#### 7) $5\sqrt{42} \times \sqrt{3}$
$$
= 5 \cdot \sqrt{42 \cdot 3} = 5\sqrt{126}
$$
Now $126 = 9 \cdot 14 = 3^2 \cdot 14$, so:
$$
= 5 \cdot \sqrt{9 \cdot 14} = 5 \cdot 3\sqrt{14} = 15\sqrt{14}
$$

#### 8) $\sqrt{3} \times -\sqrt{25}$
$\sqrt{25} = 5$, so:
$$
= \sqrt{3} \cdot (-5) = -5\sqrt{3}
$$

#### 9) $\sqrt{99} \times \sqrt{48}$
$$
= \sqrt{99 \cdot 48}
$$
But better to simplify first:
- $\sqrt{99} = \sqrt{9 \cdot 11} = 3\sqrt{11}$
- $\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}$
So:
$$
= 3\sqrt{11} \cdot 4\sqrt{3} = 12\sqrt{33}
$$

#### 10) $5\sqrt{45} \times 3\sqrt{176}$
Simplify:
- $\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}$
- $\sqrt{176} = \sqrt{16 \cdot 11} = 4\sqrt{11}$
So:
$$
= 5 \cdot 3\sqrt{5} \cdot 3 \cdot 4\sqrt{11} = 5 \cdot 3 \cdot 3 \cdot 4 \cdot \sqrt{5} \cdot \sqrt{11}
$$
Wait: $5 \cdot 3 \cdot 4 \cdot 3 = 180$, and $\sqrt{5 \cdot 11} = \sqrt{55}$, so:
$$
= 180\sqrt{55}
$$

Wait — correction:
Actually: $5 \cdot 3\sqrt{5} \cdot 3 \cdot 4\sqrt{11} = 5 \cdot 3 \cdot 3 \cdot 4 \cdot \sqrt{5} \cdot \sqrt{11} = 180\sqrt{55}$

✔ Correct.

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#### 11) $\sqrt{12}(3 + \sqrt{3})$
$\sqrt{12} = 2\sqrt{3}$, so:
$$
= 2\sqrt{3}(3 + \sqrt{3}) = 2\sqrt{3} \cdot 3 + 2\sqrt{3} \cdot \sqrt{3} = 6\sqrt{3} + 2 \cdot 3 = 6\sqrt{3} + 6
$$

#### 12) $\sqrt{23x^2} \times \sqrt{23x}$
Note: $\sqrt{23x^2} = \sqrt{23} \cdot \sqrt{x^2} = x\sqrt{23}$ (since $x \geq 0$ assumed)
And $\sqrt{23x} = \sqrt{23} \cdot \sqrt{x}$

So:
$$
= x\sqrt{23} \cdot \sqrt{23} \cdot \sqrt{x} = x \cdot 23 \cdot \sqrt{x} = 23x\sqrt{x}
$$

#### 13) $-5\sqrt{12} \times -\sqrt{3}$
First, $\sqrt{12} = 2\sqrt{3}$, so:
$$
= -5 \cdot 2\sqrt{3} \cdot (-\sqrt{3}) = -10\sqrt{3} \cdot (-\sqrt{3}) = 10 \cdot (\sqrt{3} \cdot \sqrt{3}) = 10 \cdot 3 = 30
$$

#### 14) $2\sqrt{20x^2} \times \sqrt{5x^2}$
Simplify:
- $\sqrt{20x^2} = \sqrt{4 \cdot 5 \cdot x^2} = 2x\sqrt{5}$
- $\sqrt{5x^2} = x\sqrt{5}$

So:
$$
= 2 \cdot 2x\sqrt{5} \cdot x\sqrt{5} = 4x \cdot x \cdot \sqrt{5} \cdot \sqrt{5} = 4x^2 \cdot 5 = 20x^2
$$

#### 15) $\sqrt{12x^2} \times \sqrt{2x^3}$
$\sqrt{12x^2} = \sqrt{4 \cdot 3 \cdot x^2} = 2x\sqrt{3}$
$\sqrt{2x^3} = \sqrt{2 \cdot x^2 \cdot x} = x\sqrt{2x}$

So:
$$
= 2x\sqrt{3} \cdot x\sqrt{2x} = 2x^2 \cdot \sqrt{3 \cdot 2x} = 2x^2 \sqrt{6x}
$$

#### 16) $-12\sqrt{7x} \times \sqrt{5x^3}$
$\sqrt{5x^3} = \sqrt{5 \cdot x^2 \cdot x} = x\sqrt{5x}$

So:
$$
= -12\sqrt{7x} \cdot x\sqrt{5x} = -12x \cdot \sqrt{7x \cdot 5x} = -12x \cdot \sqrt{35x^2}
$$
$\sqrt{35x^2} = x\sqrt{35}$, so:
$$
= -12x \cdot x\sqrt{35} = -12x^2\sqrt{35}
$$

#### 17) $-5\sqrt{9x^3} \times 6\sqrt{3x^2}$
Simplify:
- $\sqrt{9x^3} = \sqrt{9 \cdot x^2 \cdot x} = 3x\sqrt{x}$
- $\sqrt{3x^2} = x\sqrt{3}$

So:
$$
= -5 \cdot 3x\sqrt{x} \cdot 6 \cdot x\sqrt{3} = -5 \cdot 3 \cdot 6 \cdot x \cdot x \cdot \sqrt{x} \cdot \sqrt{3}
$$
$$
= -90x^2 \cdot \sqrt{3x}
$$

#### 18) $-2\sqrt{12}(3 + \sqrt{12})$
$\sqrt{12} = 2\sqrt{3}$, so:
$$
= -2(2\sqrt{3})(3 + 2\sqrt{3}) = -4\sqrt{3}(3 + 2\sqrt{3})
$$
Distribute:
$$
= -4\sqrt{3} \cdot 3 + (-4\sqrt{3}) \cdot 2\sqrt{3} = -12\sqrt{3} - 8 \cdot 3 = -12\sqrt{3} - 24
$$

#### 19) $\sqrt{18x}(4 - \sqrt{6x})$
$\sqrt{18x} = \sqrt{9 \cdot 2x} = 3\sqrt{2x}$

So:
$$
= 3\sqrt{2x}(4 - \sqrt{6x}) = 3\sqrt{2x} \cdot 4 - 3\sqrt{2x} \cdot \sqrt{6x}
$$
$$
= 12\sqrt{2x} - 3\sqrt{12x^2}
$$
Now $\sqrt{12x^2} = \sqrt{4 \cdot 3 \cdot x^2} = 2x\sqrt{3}$, so:
$$
= 12\sqrt{2x} - 3 \cdot 2x\sqrt{3} = 12\sqrt{2x} - 6x\sqrt{3}
$$

#### 20) $\sqrt{3x}(6\sqrt{x^3} + \sqrt{27})$
Simplify:
- $\sqrt{x^3} = x\sqrt{x}$
- $\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}$

So:
$$
= \sqrt{3x}(6x\sqrt{x} + 3\sqrt{3}) = \sqrt{3x} \cdot 6x\sqrt{x} + \sqrt{3x} \cdot 3\sqrt{3}
$$
First term:
$$
6x \cdot \sqrt{3x \cdot x} = 6x \cdot \sqrt{3x^2} = 6x \cdot x\sqrt{3} = 6x^2\sqrt{3}
$$
Second term:
$$
3 \cdot \sqrt{3x \cdot 3} = 3 \cdot \sqrt{9x} = 3 \cdot 3\sqrt{x} = 9\sqrt{x}
$$
Wait — let's do it properly:
$$
\sqrt{3x} \cdot 3\sqrt{3} = 3 \cdot \sqrt{3x \cdot 3} = 3 \cdot \sqrt{9x} = 3 \cdot 3\sqrt{x} = 9\sqrt{x}
$$

So total:
$$
= 6x^2\sqrt{3} + 9\sqrt{x}
$$

#### 21) $\sqrt{15r}(5 + \sqrt{5})$
Distribute:
$$
= \sqrt{15r} \cdot 5 + \sqrt{15r} \cdot \sqrt{5} = 5\sqrt{15r} + \sqrt{75r}
$$
Now $\sqrt{75r} = \sqrt{25 \cdot 3r} = 5\sqrt{3r}$

So:
$$
= 5\sqrt{15r} + 5\sqrt{3r}
$$

#### 22) $-5\sqrt{3x} \times 4\sqrt{6x^3}$
Simplify:
- $\sqrt{6x^3} = \sqrt{6 \cdot x^2 \cdot x} = x\sqrt{6x}$

So:
$$
= -5 \cdot 4 \cdot \sqrt{3x} \cdot x\sqrt{6x} = -20x \cdot \sqrt{3x \cdot 6x} = -20x \cdot \sqrt{18x^2}
$$
$\sqrt{18x^2} = \sqrt{9 \cdot 2 \cdot x^2} = 3x\sqrt{2}$, so:
$$
= -20x \cdot 3x\sqrt{2} = -60x^2\sqrt{2}
$$

#### 23) $-2\sqrt{18x} \times 4\sqrt{2x}$
Simplify:
- $\sqrt{18x} = \sqrt{9 \cdot 2x} = 3\sqrt{2x}$

So:
$$
= -2 \cdot 3\sqrt{2x} \cdot 4\sqrt{2x} = -6 \cdot 4 \cdot \sqrt{2x} \cdot \sqrt{2x} = -24 \cdot (2x) = -48x
$$

#### 24) $-3\sqrt{5v^2} \times (-3\sqrt{15v})$
Simplify:
- $\sqrt{5v^2} = v\sqrt{5}$

So:
$$
= -3 \cdot v\sqrt{5} \cdot (-3)\sqrt{15v} = 9v \cdot \sqrt{5} \cdot \sqrt{15v}
$$
$$
= 9v \cdot \sqrt{75v} = 9v \cdot \sqrt{25 \cdot 3v} = 9v \cdot 5\sqrt{3v} = 45v\sqrt{3v}
$$

#### 25) $(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})$
This is a difference of squares:
$$
= (\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2
$$

#### 26) $(-4\sqrt{6} + 2)(\sqrt{6} - 5)$
Use distributive property (FOIL):
$$
= -4\sqrt{6} \cdot \sqrt{6} + (-4\sqrt{6}) \cdot (-5) + 2 \cdot \sqrt{6} + 2 \cdot (-5)
$$
$$
= -4 \cdot 6 + 20\sqrt{6} + 2\sqrt{6} - 10 = -24 + 22\sqrt{6} - 10 = -34 + 22\sqrt{6}
$$

#### 27) $(2 - 2\sqrt{3})(-2 + \sqrt{3})$
FOIL:
$$
= 2 \cdot (-2) + 2 \cdot \sqrt{3} + (-2\sqrt{3}) \cdot (-2) + (-2\sqrt{3}) \cdot \sqrt{3}
$$
$$
= -4 + 2\sqrt{3} + 4\sqrt{3} - 2 \cdot 3 = -4 + 6\sqrt{3} - 6 = -10 + 6\sqrt{3}
$$

#### 28) $(11 - 4\sqrt{5})(6 - \sqrt{5})$
FOIL:
$$
= 11 \cdot 6 + 11 \cdot (-\sqrt{5}) + (-4\sqrt{5}) \cdot 6 + (-4\sqrt{5})(-\sqrt{5})
$$
$$
= 66 - 11\sqrt{5} - 24\sqrt{5} + 4 \cdot 5 = 66 - 35\sqrt{5} + 20 = 86 - 35\sqrt{5}
$$

#### 29) $(-2 - \sqrt{3x})(3 + \sqrt{3x})$
FOIL:
$$
= -2 \cdot 3 + (-2)(\sqrt{3x}) + (-\sqrt{3x})(3) + (-\sqrt{3x})(\sqrt{3x})
$$
$$
= -6 -2\sqrt{3x} -3\sqrt{3x} - (3x) = -6 -5\sqrt{3x} -3x
$$
$$
= -3x -6 -5\sqrt{3x}
$$

#### 30) $(-2 + 3\sqrt{2r})(-2 + \sqrt{2r})$
FOIL:
$$
= (-2)(-2) + (-2)(\sqrt{2r}) + (3\sqrt{2r})(-2) + (3\sqrt{2r})(\sqrt{2r})
$$
$$
= 4 -2\sqrt{2r} -6\sqrt{2r} + 3 \cdot 2r = 4 -8\sqrt{2r} + 6r
$$
$$
= 6r + 4 -8\sqrt{2r}
$$

#### 31) $(-4\sqrt{2n} + 2)(-2\sqrt{2} - 4)$
FOIL:
$$
= (-4\sqrt{2n})(-2\sqrt{2}) + (-4\sqrt{2n})(-4) + 2(-2\sqrt{2}) + 2(-4)
$$
Compute each:
- $(-4\sqrt{2n})(-2\sqrt{2}) = 8 \cdot \sqrt{2n \cdot 2} = 8 \cdot \sqrt{4n} = 8 \cdot 2\sqrt{n} = 16\sqrt{n}$
- $(-4\sqrt{2n})(-4) = 16\sqrt{2n}$
- $2(-2\sqrt{2}) = -4\sqrt{2}$
- $2(-4) = -8$

So total:
$$
= 16\sqrt{n} + 16\sqrt{2n} -4\sqrt{2} -8
$$

#### 32) $(-1 + 2\sqrt{3})(2 - 3\sqrt{3x})$
FOIL:
$$
= (-1)(2) + (-1)(-3\sqrt{3x}) + (2\sqrt{3})(2) + (2\sqrt{3})(-3\sqrt{3x})
$$
$$
= -2 + 3\sqrt{3x} + 4\sqrt{3} -6\sqrt{3} \cdot \sqrt{3x}
$$
Now $\sqrt{3} \cdot \sqrt{3x} = \sqrt{9x} = 3\sqrt{x}$, so:
$$
-6 \cdot 3\sqrt{x} = -18\sqrt{x}
$$
So:
$$
= -2 + 3\sqrt{3x} + 4\sqrt{3} -18\sqrt{x}
$$

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $5$ |
| 2 | $5\sqrt{2}$ |
| 3 | $6$ |
| 4 | $7\sqrt{6}$ |
| 5 | $-40$ |
| 6 | $18\sqrt{2}$ |
| 7 | $15\sqrt{14}$ |
| 8 | $-5\sqrt{3}$ |
| 9 | $12\sqrt{33}$ |
| 10 | $180\sqrt{55}$ |
| 11 | $6 + 6\sqrt{3}$ |
| 12 | $23x\sqrt{x}$ |
| 13 | $30$ |
| 14 | $20x^2$ |
| 15 | $2x^2\sqrt{6x}$ |
| 16 | $-12x^2\sqrt{35}$ |
| 17 | $-90x^2\sqrt{3x}$ |
| 18 | $-24 - 12\sqrt{3}$ |
| 19 | $12\sqrt{2x} - 6x\sqrt{3}$ |
| 20 | $6x^2\sqrt{3} + 9\sqrt{x}$ |
| 21 | $5\sqrt{15r} + 5\sqrt{3r}$ |
| 22 | $-60x^2\sqrt{2}$ |
| 23 | $-48x$ |
| 24 | $45v\sqrt{3v}$ |
| 25 | $2$ |
| 26 | $-34 + 22\sqrt{6}$ |
| 27 | $-10 + 6\sqrt{3}$ |
| 28 | $86 - 35\sqrt{5}$ |
| 29 | $-3x -6 -5\sqrt{3x}$ |
| 30 | $6r + 4 -8\sqrt{2r}$ |
| 31 | $16\sqrt{n} + 16\sqrt{2n} -4\sqrt{2} -8$ |
| 32 | $-2 + 3\sqrt{3x} + 4\sqrt{3} -18\sqrt{x}$ |

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