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anonymous
 3 years ago
Factor the trinomial: 4x^2  8x  5
(2x + 1)(2x  5)
(4x + 1)(x  5)
(4x  5)(x + 1)
prime
anonymous
 3 years ago
Factor the trinomial: 4x^2  8x  5 (2x + 1)(2x  5) (4x + 1)(x  5) (4x  5)(x + 1) prime

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1trying to pick from a multiple choice is simpler than trying to work thru the math. I always separate taking multiple choice from actually doing math.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1we can deduce the the FL parts of foil are the same for every given option. if we can reduce this to the proper OI part that would reduce the work load

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im not trying to find the answer im trying to understand it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know that the first thing i have to do it multiply the first and last number right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1now multiply the "inside" numbers

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0um i thought i was suppose to find the factors of 20 then use that factors that add to 8?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1im not sure if we are seeing the same thing at the moment. take the forms: (a+b)(c+d) if ac equals the first term; and bd equals the last term; then we need to determine the middle term by: ac+bd

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1i know its not this one, but to example my idea ... (4x  5)(x + 1) 4x*x = 4x^2 5*1 = 5 and our given setup is 4x^2 8x 5 all the options give us the same first and last terms so we need to determine a suitable "middle" setup

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im pretty sure im not doing foil im ffactoring the trinomial

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1(4x  5)(x + 1) 4x * 1 = 4x 5 * x = 5x 4x  5x is not equal to 8x so we can rule that out

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1factoring produces a foil; and foil results in the equation they want factored 10,2 is perfect for us

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so id write it like thsi? 4x^2+2x10x5?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1that looks suitable to me; i was never really good at the factor by grouping stuff :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so then what do i do thats no9t one of te answers?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1then you group it, and factor our the groups, then factor again ... which is why i never really go that route in these ... (4x^2+2x)+(10x5) 2x(2x+1)5(2x+1) (2x+1)(2x5)

rosedewittbukater
 3 years ago
Best ResponseYou've already chosen the best response.1I responded to your other question of this.

rosedewittbukater
 3 years ago
Best ResponseYou've already chosen the best response.14x^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. \[4x ^{2}+2x10x5\]Then group the equation into 2 separate parts. \[(4x ^{2}+2x)+(10x5)\]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. same.\[(2x5)(2x+1)\]It becomes this because 2x+1 are the There's your answer. Hope this helps!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thank you all soooooooooo much!!!!

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1rose has a pretty detailed answer; not alot of interaction allowed in it, but a pretty detailed answer nonetheless :)
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