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Multiplying Binomials

Module by: Kenny M. Felder. E-mail the author

Summary: Some practice multiplying binomials to get quadratic equations.

(*or, "These are a few of my favorite formulae)

Exercise 1

Multiply: (x+2)(x+2)(x+2)(x+2) size 12{ \( x+2 \) \( x+2 \) } {}

Test your result by plugging x=3x=3 size 12{x=3} {} into both my original function, and your resultant function. Do they come out the same?

Exercise 2

Multiply: (x+3)(x+3)(x+3)(x+3) size 12{ \( x+3 \) \( x+3 \) } {}

Test your result by plugging x=1x=1 size 12{x= - 1} {} into both my original function, and your resultant function. Do they come out the same?

Exercise 3

Multiply: (x+5)(x+5)(x+5)(x+5) size 12{ \( x+5 \) \( x+5 \) } {}

Test your result by plugging x=x= size 12{x= { size 8{1} } wideslash { size 8{2} } } {} into both my original function, and your resultant function. Do they come out the same?

Exercise 4

Multiply: (x+a)(x+a)(x+a)(x+a) size 12{ \( x+a \) \( x+a \) } {}

Now, leave xx size 12{x} {} as it is, but plug a=3a=3 size 12{a=3} {} into both my original function, and your resultant function. Do you get two functions that are equal? Do they look familiar?

Exercise 5

Do not answer these questions by multiplying them out explicitly. Instead, plug these numbers into the general formula for (x+a)2(x+a)2 size 12{ \( x+a \) rSup { size 8{2} } } {} that you found in number 4.

  • a. ( x + 4 ) ( x + 4 ) ( x + 4 ) ( x + 4 ) size 12{ \( x+4 \) \( x+4 \) } {}
  • b. ( y + 7 ) 2 ( y + 7 ) 2 size 12{ \( y+7 \) rSup { size 8{2} } } {}
  • c. ( z + ) 2 ( z + ) 2 size 12{ \( z+ { size 8{1} } wideslash { size 8{2} } \) rSup { size 8{2} } } {}
  • d. ( m + 2 ) 2 ( m + 2 ) 2 size 12{ \( m+ sqrt {2} \) rSup { size 8{2} } } {}
  • e. (x3)2(x3)2 size 12{ \( x - 3 \) rSup { size 8{2} } } {} (*so in this case, (a(a size 12{ \( a} {} is –3.)
  • f. ( x 1 ) 2 ( x 1 ) 2 size 12{ \( x - 1 \) rSup { size 8{2} } } {}
  • g. ( x a ) 2 ( x a ) 2 size 12{ \( x - a \) rSup { size 8{2} } } {}

Exercise 6

Earlier in class, we found the following generalization: (x+a)(xa)=x2a2(x+a)(xa)=x2a2 size 12{ \( x+a \) \( x - a \) =x rSup { size 8{2} } - a rSup { size 8{2} } } {}. Just to refresh your memory on how we found that, test this generalization for the following cases.

  • a. x = 10 , a = 0 x = 10 , a = 0 size 12{x="10"", "a=0} {}
  • b. x = 10 , a = 1 x = 10 , a = 1 size 12{x="10"", "a=1} {}
  • c. x = 10 , a = 2 x = 10 , a = 2 size 12{x="10"", "a=2} {}
  • d. x = 10 , a = 3 x = 10 , a = 3 size 12{x="10"", "a=3} {}

Exercise 7

Test the same generalization by multiplying out (x+a)(xa)(x+a)(xa) size 12{ \( x+a \) \( x - a \) } {} explicitly.

Exercise 8

Now, use that “difference between two squares” generalization. As in #5, do not solve these by multiplying them out, but by plugging appropriate values into the generalization in #6.

  • a. ( 20 + 1 ) ( 20 1 ) = ( 20 + 1 ) ( 20 1 ) = size 12{ \( "20"+1 \) \( "20" - 1 \) ={}} {}
  • b. ( x + 3 ) ( x 3 ) = ( x + 3 ) ( x 3 ) = size 12{ \( x+3 \) \( x - 3 \) ={}} {}
  • c. ( x + 2 ) ( x 2 ) ( x + 2 ) ( x 2 ) size 12{ \( x+ sqrt {2} \) \( x - sqrt {2} \) } {}
  • d. ( x + 2 / 3 ) ( x 2 / 3 ) ( x + 2 / 3 ) ( x 2 / 3 ) size 12{ \( x+ {2} slash {3} \) \( x - {2} slash {3} \) } {}

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