Multiplying Radical Expressions

Multiplying Radical Expressions exercise

• Explain how to multiply $(15\sqrt5+5\sqrt3)(6\sqrt3-2\sqrt5)$.

  Hints

  FOIL is as way of remembering a method of multiplying two binomials:

  • Multiply the First.
  • Multiply the Outer.
  • Multiply the Inner.
  • Multiply the Last.

  Here is an example of multiplication using the FOIL method:

  $(3+b)(2+b)=6+3b+2b+b^2$

  Now you can combine like terms to get:

  $6+3b+2b+b^2=6+5b+b^2$

  Solution

  To multiply $(15\sqrt5+5\sqrt3)(6\sqrt3-2\sqrt5)$, first use the FOIL method to multiply these binomials:

  • F multiply the first $15\sqrt5(6\sqrt3)$
  • O multiply the outer $-15\sqrt5(2\sqrt5)$
  • I multiply the inner $5\sqrt3(6\sqrt3)$
  • L multiply the last $-5\sqrt3(2\sqrt5)$

  Adding these terms together we get

  $(15\sqrt5+5\sqrt3)(6\sqrt3-2\sqrt5)=15\sqrt5(6\sqrt3)-15\sqrt5(2\sqrt5)+5\sqrt3(6\sqrt3)-5\sqrt3(2\sqrt5)$.

  Multiply the radicals and combine like terms:

  $\begin{array}{rcl} (15\sqrt5+5\sqrt3)(6\sqrt3-2\sqrt5)&=&~90\sqrt{15}-30\sqrt{25}+30\sqrt9-10\sqrt{15}\\ &=&~90\sqrt{15}-10\sqrt{15}-150+90\\ &=&~80\sqrt{15}-60 \end{array}$

  There is no further simplification possible, so we have our answer.

• Calculate $(15\sqrt5+5\sqrt3)(6\sqrt5-2\sqrt3)$.

  Hints

  Here is an example for using the FOIL method:

  $\begin{array}{rcl} (3+a)(2+2a)&=&~6+3(2a)+2a+a(2a)\\ &=&~6+6a+2a+2a^2\\ &=&~2a^2+8a+6 \end{array}$

  Keep in mind that the square root of a square is equal to the term being squared:

  $\sqrt{a^2}=a$.

  The square root of a product is the product of the square roots of its factors.

  Solution

  To multiply $(15\sqrt5+5\sqrt3)(6\sqrt5-2\sqrt3)$, first use the FOIL method to multiply these binomials:

  • F multiply the first $15\sqrt5(6\sqrt5)$
  • O multiply the outer $-15\sqrt5(2\sqrt3)$
  • I multiply the inner $5\sqrt3(6\sqrt5)$
  • L multiply the last $-5\sqrt3(2\sqrt3)$

  Now we add the terms to get

  $(15\sqrt5+5\sqrt3)(6\sqrt5-2\sqrt3)=15\sqrt5(6\sqrt5)-15\sqrt5(2\sqrt3)+5\sqrt3(6\sqrt5)-5\sqrt3(2\sqrt3)$.

  We multiply the radicals and combine the like terms to get:

  $(15\sqrt5+5\sqrt3)(6\sqrt5-2\sqrt3)=90\sqrt{25}-30\sqrt{15}+30\sqrt{15}-10\sqrt9$.

  Now we can simplify the radicals and cancel out the term $-30\sqrt{15}$ with $30\sqrt{15}$, giving us that

  $(15\sqrt5+5\sqrt3)(6\sqrt5-2\sqrt3)=450-30=420$.

• Identify the steps of multiplication.

  Hints

  • F multiply the first $3\sqrt3(4\sqrt3)$
  • O multiply the outer $-3\sqrt3(2\sqrt2)$
  • I multiply the inner $5\sqrt2(4\sqrt3)$
  • L multiply the last $-5\sqrt2(2\sqrt2)$

  Here you see an example using the FOIL method.

  The second step in the example above is combining like terms:

  $5a-2a=3a$.

  Solution

  Using the FOIL method we get

  $(3\sqrt3+5\sqrt2)(4\sqrt3-2\sqrt2)=3\sqrt3(4\sqrt3)-3\sqrt3(2\sqrt2)+5\sqrt2(4\sqrt3)-5\sqrt2(2\sqrt2)$.

  We multiply the radicals,

  $(3\sqrt3+5\sqrt2)(4\sqrt3-2\sqrt2)=12\sqrt9-6\sqrt6+20\sqrt6-10\sqrt4$,

  and combine the like terms,

  $(3\sqrt3+5\sqrt2)(4\sqrt3-2\sqrt2)=12\sqrt9+14\sqrt6-10\sqrt4$.

  Now we calculate the square roots of squares:

  • $\sqrt9=\sqrt{3^2}=3$
  • $\sqrt4=\sqrt{2^2}=2$

  $(3\sqrt3+5\sqrt2)(4\sqrt3-2\sqrt2)=12(3)+14\sqrt6-10(2)$.

  And finally we combine like terms again to get

  $(3\sqrt3+5\sqrt2)(4\sqrt3-2\sqrt2)=36+14\sqrt6-20=16+14\sqrt6$.

• Calculate the product.

  Hints

  • F multiply the first $12\sqrt6(12\sqrt6)$
  • O multiply the outer $12\sqrt6(3\sqrt5)$
  • I multiply the inner $-3\sqrt5(12\sqrt6)$
  • L multiply the last $-3\sqrt5(3\sqrt5)$

  You still have to add the terms calculated using the FOIL method together.

  Keep the following in mind that $\sqrt{a^2}=a$. For example:

  Solution

  We use the FOIL method first to get

  $(12\sqrt6-3\sqrt5)(12\sqrt6+3\sqrt5)=12\sqrt6(12\sqrt6)+12\sqrt6(3\sqrt5)-3\sqrt6(12\sqrt6)-3\sqrt5(3\sqrt5)$.

  Next we multiply the radicals and combine like terms:

  $(12\sqrt6-3\sqrt5)(12\sqrt6+3\sqrt5)=144\sqrt{36}+36\sqrt{30}-36\sqrt{30}-9\sqrt{25}$.

  Once again, combining like terms and calculating the radicals of squares we get:

  $(12\sqrt6-3\sqrt5)(12\sqrt6+3\sqrt5)=144(6)-9(5)=864-45=819$.

• Decide which expressions can be simplified.

  Hints

  You can only combine like terms.

  For example, $5\sqrt 3+6\sqrt3=11\sqrt3$, but you can't combine $5\sqrt 3+6\sqrt5$.

  For example:

  $\begin{array}{rcl} 2\sqrt5(5\sqrt3)&=&~2\times 5\times\sqrt5\times \sqrt3\\ &=&~10\times\sqrt{5\times 3}\\ &=&~10\times\sqrt{15} \end{array}$

  There are just three terms which can be simplified.

  Solution

  Combining like terms means that you only can add or subtract terms with the same variable and exponent:

  • $a+a=2a$
  • $a^2+a^2=2a^2$

  But $a+a^2$ can't be simplified any further, for instance.

  Let's have a look at the following example:

  $(15\sqrt5+5\sqrt3)(3\sqrt2-5\sqrt6)$.

  First we use the FOIL method:

  • F multiply the first $15\sqrt5(3\sqrt2)$
  • O multiply the outer $-15\sqrt5(5\sqrt6)$
  • I multiply the inner $5\sqrt3(3\sqrt2)$
  • L multiply the last $-5\sqrt3(5\sqrt6)$

  Adding all of these resulting terms together, we get:

  $(15\sqrt5+5\sqrt3)(3\sqrt2-5\sqrt6)=15\sqrt5(3\sqrt2)-15\sqrt5(5\sqrt6)+5\sqrt3(3\sqrt2)-5\sqrt3(5\sqrt6)$.

  The radicals can be multiplied by multiplying the terms under the square roots:

  • $\sqrt5(\sqrt2)=\sqrt{10}$
  • $\sqrt5(\sqrt6)=\sqrt{30}$
  • $\sqrt3(\sqrt2)=\sqrt{6}$
  • $\sqrt3(\sqrt6)=\sqrt{18}$

  So we get $(15\sqrt5+5\sqrt3)(3\sqrt2-5\sqrt6)=45\sqrt{10}-75\sqrt{30}+15\sqrt6-25\sqrt{18}$.

  We are still able to simplify $\sqrt{18}=\sqrt{9\times 2}=\sqrt9\times\sqrt2=3\sqrt2$ further.

  This together with the expression above gives us:

  $(15\sqrt5+5\sqrt3)(3\sqrt2-5\sqrt6)=45\sqrt{10}-75\sqrt{30}+15\sqrt6-75\sqrt{2}$

  And that's it! We have simplified the expression as much as we can.

• Find the errors in the calculation.

  Hints

  Every calculation has an error.

  The result of the first multiplication is $-7\sqrt{10}+3$.

  Keep in mind that you can only combine like terms.

  Here is an example for multiplying radicals:

  $\begin{array}{rcl} \sqrt 3\times \sqrt2&=&~\sqrt{3\times 2}=\sqrt6\\\\ \sqrt{18}&=&~\sqrt{9\times 2}=\sqrt9\times \sqrt 2=3\sqrt2 \end{array}$

  Solution

  Let's first outline the steps for multiplying expressions with radicals:

  1. Use the FOIL method.
  2. Rearrange the terms and combine like terms. Remember that $\sqrt a\times \sqrt b=\sqrt{a\times b}$.
  3. Calculate the radicals if it's possible; for example $\sqrt 4=\sqrt{2^2}=2$.
  4. Simplify the result as much as possible by combining like terms.

  Let's practice by going over the examples presented in the problem:

  $~$

  For $(\sqrt5-3\sqrt2)(2\sqrt2+3\sqrt5)$:

  First use the FOIL method:

  $(\sqrt5-3\sqrt2)(2\sqrt2+3\sqrt5)=\sqrt5(2\sqrt2)+\sqrt5(3\sqrt5)-3\sqrt2(2\sqrt2)-3\sqrt2(3\sqrt5)$.

  Second rearrange the terms and combine like terms:

  $\begin{array}{rcl} ((\sqrt5-3\sqrt2)(2\sqrt2+3\sqrt5)&=&~2\sqrt{10}+3\sqrt{25}-6\sqrt4-9\sqrt{10}\\ &=&~2\sqrt{10}-9\sqrt{10}+3\sqrt{25}-6\sqrt4\\ &=&~-7\sqrt{10}+3\sqrt{25}-6\sqrt4. \end{array}$

  Third calculate the radicals if possible: $(\sqrt5-3\sqrt2)(2\sqrt2+3\sqrt5)=-7\sqrt{10}+3(5)-6(2)$.

  Fourth simplify the result as much as possible: $(\sqrt5-3\sqrt2)(2\sqrt2+3\sqrt5) =-7\sqrt{10}+15-12=-7\sqrt{10}+3$.

  $~$

  For $(2\sqrt6+3\sqrt5)(3\sqrt6-5\sqrt3)$:

  1. $(2\sqrt6+3\sqrt5)(3\sqrt6-5\sqrt3)=2\sqrt6(3\sqrt6)-2\sqrt6(5\sqrt3)+3\sqrt5(3\sqrt6)-3\sqrt5(5\sqrt3)$
  2. $(2\sqrt6+3\sqrt5)(3\sqrt6-5\sqrt3)=6\sqrt{36}-10\sqrt{18}+9\sqrt{30}-15\sqrt{15}=6\sqrt{36}-30\sqrt{2}+9\sqrt{30}-15\sqrt{15}$
  3. $(2\sqrt6+3\sqrt5)(3\sqrt6-5\sqrt3)=6(6)-30\sqrt{2}+9\sqrt{30}-15\sqrt{15}=36-30\sqrt{2}+9\sqrt{30}-15\sqrt{15}$
  4. There is no further simplification possible.