Factoring Special Case Polynomials

Factoring Special Case Polynomials exercise

• Determine the area of the pool.

  Hints

  If you've got any square with the side $s$, then the area is given by $s\times s=s^2$.

  Use FOIL for multiplying two binomials. For example, let's factor $(a+2)\times (a+2)$:

  First - multiply the first $a\times a=a^2$

  Outer - multiply the outer $a\times 2=2a$

  Inner - multiply the inner $2\times a=2a$

  Last - multiply the last $2\times 2=4$

  Adding all of the results together gives us $(a+2)^2=a^2+4a+4$.

  You can use $(a+b)(a-b)$ to compute numbers; for example:

  $102\times 98=(100+2)(100-2)=100^2-4=9996$.

  Solution

  We know the side lengths for each pool Mrs. Feinberg can choose from. We need to figure out the area of each pool so that she can make the most educated decision about which pool she wants for her epic pool party.

  $~$

  Any square with the side lengths $a$ has the area $a\times a=a^2$. With this fact and the FOIL method for multiplication, we can figure these areas out!

  Lets recall the FOIL method for multiplication:

  First - multiply the first

  Outer - multiply the outer

  Inner - multiply the inner

  Last - multiply the last

  $~$

  For the the leftmost pool, the sides are $a\times a=a^2$.

  $~$

  For the second pool, we need to multiply $(a-b)\times(a-b)$ to find the area.

  Using the FOIL method, we have:

  (1) "multiply the first" to get $a^2$.

  (2) "multiply the outer" to get $-ab$.

  (3) "multiply the inner" to get $-ab$.

  (4) "multiply the last" to get $b^2$.

  Adding all of these together gives us our area: $(a-b)\times(a-b)=a^2-2ab+b^2$.

  $~$

  Using the FOIL method in the same way for the third pool we get that the area is $(a+b)\times(a+b)=a^2+2ab+b^2$.

  $~$

  For the fourth pool, we get $(a+b)\times(a-b)=a^2-b^2$.

• Calculate the area of each pool using the given values for $a$ and $b$.

  Hints

  First calculate the the quantity inside the parenthesis and then multiply.

  Here is an example.

  Remember: PEMDAS; Parenthesis first and then Exponents.

  Solution

  From the leftmost pool to the rightmost:

  First Pool:

  • This pool is a square with side lengths $a=10~ft$.
  • So we have to square $a$ to $(10~ft)^2=100~ft^2$.

  Second Pool:

  • This pool is a square with side lengths $a-b$.
  • So first we calculate the difference, $10~ft-5~ft=5~ft$.
  • Then we square the difference $5~ft$ to get $(5~ft)^2=25~ft^2$.

  Third Pool:

  • This pool is a square with side lengths $a+b$.
  • So we calculate the sum $10~ft+5~ft=15~ft$.
  • Then we square it to get $(15~ft)^2=225~ft^2$.

  Fourth Pool:

  • This last pool is a rectangle with side lengths $10~ft+5~ft=15~ft$ and $10~ft-5~ft=5~ft$.
  • To get the corresponding area we multiply these values to get $15~ft\times 5~ft=75~ft$.

• Calculate the different possible sizes of the rose bed.

  Hints

  Calculate the quantity inside the parenthesis first.

  You'll use the following expressions:

  • $(a+b)^2$

  • $(a-b)^2$

  • $(a+b)(a-b)$

  Solution

  (1) The square rose bed with side lengths $a$ has area $a\times a=a^2=(15)^2=225~ft^2$.

  (2) The square rose bed with side lengths $a+7$ has area $(a+7)^2=(15+7)^2=22^2=484~ft^2$.

  (3) The square rose bed with side lengths $a-7$ has area $(a-7)^2=(15-7)^2=8^2=64~ft^2$.

  (4) A rectangular rose bed with one side length of $a+3$ and another side length of $a-3$ has area $(a+3)(a-3)=(15+3)(15-3)=18\times12=216~ft^2$.

• Find the carpet sizes using the FOIL method.

  Hints

  First write down the different binomial products for the area of each carpet.

  After writing down all the binomial products, calculate the quantities inside the parenthesis for each one.

  Solution

  Let's start with the smallest carpet with side lengths $12~ft+3~ft$ and $12~ft-3~ft$.

  The size of this carpet is $15\times 9=135~ft^2$.

  $~$

  Next we have the square carpet with side lengths $16~ft-3~ft$.

  The size of this carpet is $13^2=169~ft^2$.

  $~$

  The rectangular carpet with side lengths $15~ft+1~ft$ and $15~ft-1~ft$ is

  $16\times 14=224~ft^2$.

  $~$

  The square carpet with the side lengths $12~ft$ is

  $15^2=225~ft^2$.

  $~$

  The biggest carpet is the rectangular one with side lengths $16~ft+3~ft$ and $16~ft-3~ft$.

  So it is $19\times 13=247~ft^2$.

• Explain the FOIL method for multiplication.

  Hints

  The F step using the FOIL method for $(a+b)(a+b)$ is $a\times a=a^2$.

  The I step using the FOIL method for $(a+b)(a+b)$ is $b\times a=ab$.

  Solution

  To explain FOIL method for multiplication, let's have a look at the following example:

  To multiply the binomials $(a+2)(b+3)$,

  • First - multiply the first $a\times b=ab$

  • Outer - multiply the outer $a\times 3=3a$

  • Inner - multiply the inner $2\times b=2b$

  • Last - multiply the last $2\times 3=6$

  Adding all the results leads to $(a+2)(b+3)=ab+3a+2b+6$.

• Use the product of binomials to calculate $43\times 37$.

  Hints

  Use one of the equations:

  • $(a+b)^2=a^2+2ab+b^2$

  • $(a-b)^2=a^2-2ab+b^2$

  • $(a+b)(a-b)=a^2-b^2$

  Decide what the values of $a$ and $b$ need to be.

  We have to solve to equations $a+b=43$ and $a-b=37$.

  Solution

  We can use the formula $(a+b)(a-b)=a^2-b^2$ to help us solve $43\times 37$.

  $~$

  With this formula, we know that we have to solve the equations $a+b=43$ and $a-b=37$.

  Adding these equations together, we get $2a=80$.

  Then dividing by $2$ gives us $a=40$, and we can see that $b=3$.

  Let's check: $(a+b)(a-b)=(40+3)(40-3)=43\times 37$ $\surd$

  $~$

  Looking at the right-hand side of the equation, we have $a^2-b^2$.

  Substituting $a=40$ and $b=3$, we get $240^2-3^2=1600-9=1591$.

  $~$

  With this trick, multiplying numbers can seem like magic! Keep this in mind for the next time you want to impress your friends with your mental math skills.