anonymous
  • anonymous
Choose the correct product of (5x + 4)2. 25x2 + 40x + 16 25x2 − 16 25x2 + 16 25x2 − 40x + 16
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
(5x+4)(5x+4) Do you know how to use the FOIL method?
anonymous
  • anonymous
no
anonymous
  • anonymous
First, Outer, Inner, Last. Your first terms that you need to multiply are 5x and 5x 5x*5x = 25x^2 Your outer terms that you need to multiply are 5x and 4. 5x*4 = 20x Your inner terms that you need to multiply are 5x and 4. 5x*4 = 20x You last terms that you need to multiply are 4 and 4. 4*4 = 16 So you have 25x^2, 20x, 20x, and 16 that you need to "combine." 25x^2 +20x +20x +16 = 25x^2 +40x +16

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anonymous
  • anonymous
ah wow thanks so much!
anonymous
  • anonymous
Watch this :-) http://www.youtube.com/watch?v=Twtdsp0a1Iw
anonymous
  • anonymous
@phi @CrazyMexican816
anonymous
  • anonymous
is this correct
anonymous
  • anonymous
i still dont nderstand @phi
phi
  • phi
yes
anonymous
  • anonymous
i dont know anything?
phi
  • phi
see Khan's vide if you have time http://www.khanacademy.org/math/algebra/multiplying-factoring-expression/multiplying-binomials/v/multiplying-binomials
anonymous
  • anonymous
i know that \(\color{lime}{\huge\ (5x+4)^2~}\) = 25x^2+16 but i dont understand if i add a 40x
anonymous
  • anonymous
@phi ?
anonymous
  • anonymous
in FOIL i use 5x^2+4^2 but dont get why did they add a 40x?
phi
  • phi
do you know what you get when you multiply a(x+y)
anonymous
  • anonymous
ax+ay?
phi
  • phi
yes. now what if a was "complicated" in other words, if a was short-hand for (c+d) ? we would get (c+d)x + (c+d)y (I just replaced "a" with its equivalent (c+d) ) now we do the "distributive property" again: (c+d)x is the same as x(c+d), or, xc +xd and (c+d)y is cy + dy
anonymous
  • anonymous
ohh
phi
  • phi
in other words (c+d)(x+y) can be written as xc + xd +yc +yd we get FOUR terms (which would come from using FOIL, if you learn that)
anonymous
  • anonymous
foil is first out inner last
phi
  • phi
with (5x+4)(5x+4) we will get 4 terms, but we can combine two of the terms, and get 3 terms in the end
anonymous
  • anonymous
BTW i hope you dont ge confused cuz it looks like (5x + 4)•2 but it is (5x + 4)^2
phi
  • phi
let's use the distributive idea to do (5x+4)(5x+4) First, let a= (5x+4) so it's easier to see what we are doing: a(5x+4)= a*5x +a*4 now replace a with (5x+4) (5x+4)*5x + (5x+4)*4 now expand both terms: can you do that ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
5x(5x+4)+4(5x+4) 25x^2•20x+20x+16 ohhhhhh thx @phi
phi
  • phi
the first part 5x(5x+4) is written as 25x^2 +20x so you get all together \[ 25x^2 +20x +20x +16\] the last step is notice we have 20 x's plus 20 more x's which is 40 x's. we write \[ 25x^2 +40x +16\]
some.random.cool.kid
  • some.random.cool.kid
hi

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