anonymous
  • anonymous
PLEASE HELKP ME IM BEHIND PLEASE EXPLAIN STEPS AND I WILL GIVE YOU A MEDAL. Factor the trinomial: 4x^2 - 8x - 5 (2x + 1)(2x - 5) (4x + 1)(x - 5) (4x - 5)(x + 1) prime
Mathematics
schrodinger
  • schrodinger
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precal
  • precal
|dw:1360095374158:dw|
precal
  • precal
I am going to teach you a method called slide and divide take the 1st number and slide it to the back by multiplying it to the last number
anonymous
  • anonymous
so -20

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precal
  • precal
|dw:1360095443097:dw|
precal
  • precal
we know x times x is x to the second power we are using 10 and 2 because 10 times 2 is 20 but we need to subtract them to get 8 the minus sign in front of 20 tells us to subtract those two numbers
precal
  • precal
|dw:1360095532927:dw|
anonymous
  • anonymous
|dw:1360095517633:dw| what happened there?
precal
  • precal
+2-10 gives us -8 we need -8x
precal
  • precal
last step is to divid the number back in
precal
  • precal
|dw:1360095575598:dw|
precal
  • precal
because 1/2 is a decimal we will now take the 2 and move it in front of x and drop the 1
precal
  • precal
|dw:1360095620919:dw|we are done
anonymous
  • anonymous
hey please slow down
precal
  • precal
you can read through each step until you learn the pattern
anonymous
  • anonymous
|dw:1360095653585:dw| you lost me here
anonymous
  • anonymous
i dont understand what you did
precal
  • precal
|dw:1360095676993:dw|you have to list all of the products of 20
precal
  • precal
the sign in front of 20 tells you to subtract them, you are looking for the middle term
precal
  • precal
|dw:1360095746738:dw|
precal
  • precal
|dw:1360095763812:dw|use 2 and 10
anonymous
  • anonymous
but that would be positive 20
precal
  • precal
|dw:1360095780117:dw|these are your sign choices
precal
  • precal
do the signs separate
precal
  • precal
|dw:1360095836253:dw|
anonymous
  • anonymous
ok look i found -2, and 10 to be the correct factors all i wanted to know was how to inser that into the equation and solve
precal
  • precal
|dw:1360095897773:dw|
precal
  • precal
2 must be positive and 10 must be negative
precal
  • precal
|dw:1360095963415:dw|
anonymous
  • anonymous
ok im sorry i made a huge mistaje when writting this i didnt use the ^
anonymous
  • anonymous
look at the prob again
precal
  • precal
does not change my answer since I did use the correct notation
anonymous
  • anonymous
can you please not use the drawing i cant read it or understand it..im sorry im just so confused at how you got your answer and i understand if you want to just give up on me..
precal
  • precal
sorry need to go, repost your problem someone will come help you
anonymous
  • anonymous
alright
rosedewittbukater
  • rosedewittbukater
4x^2 - 8x - 5 Ok so I'll try to explain this the way I understand it. Usually to factor you have to find 2 numbers that multiply to equal the last number, and add up to equal the number part of the middle. I'll just use \[ax ^{2}+bx+c\] as an example equation. So you have to find 2 numbers that multiply to get c, and add up to get b. Then you would usually take those 2 numbers and use it to factor and get (x+?)(x+?). But since there is a number for a that is not 1 in your equation, it's a little different. First, multiply the first and last numbers 4 and -5 (a and c in the example equation). You get -20. So try to find 2 number that are factors of -20 and add up to equal 8. Just start to list the factors off on paper or in your head. Remember to include the negatives. -1, 20 1, -20 -2, 10 2, -10 These add up to equal -8 Then put 2 and -10 to separate the middle part of the equation. \[4x ^{2}+2x-10x-5\]Then group the equation into 2 separate parts. \[(4x ^{2}+2x)+(-10x-5)\]Then you take out any common factors from the parentheses.\[2x(2x+1)-5(2x+1)\]And then group together the parts you took out, which are the 2x and -5.\[(2x-5)(2x+1)\]It becomes this because 2x+1 are the same. There's your answer. Hope this helps!
precal
  • precal
yes rose is showing the method known as factor by grouping. Both methods are correct

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