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Factor the trinomial: 4x^2 - 8x - 5 (2x + 1)(2x - 5) (4x + 1)(x - 5) (4x - 5)(x + 1) prime

Mathematics
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how do i solve this?
trying 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.
we 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

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Other answers:

im not trying to find the answer im trying to understand it
i know that the first thing i have to do it multiply the first and last number right?
which would be -20
now multiply the "inside" numbers
um i thought i was suppose to find the factors of -20 then use that factors that add to -8?
im 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
i 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
im pretty sure im not doing foil im ffactoring the trinomial
(4x - 5)(x + 1) 4x * 1 = 4x -5 * x = -5x 4x - 5x is not equal to -8x so we can rule that out
ok so its -10,2?
factoring produces a foil; and foil results in the equation they want factored -10,2 is perfect for us
so id write it like thsi? 4x^2+2x-10x-5?
that looks suitable to me; i was never really good at the factor by grouping stuff :)
so then what do i do thats no9t one of te answers?
then 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)+(-10x-5) 2x(2x+1)-5(2x+1) (2x+1)(2x-5)
I responded to your other question of this.
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. same.\[(2x-5)(2x+1)\]It becomes this because 2x+1 are the There's your answer. Hope this helps!
thank you all soooooooooo much!!!!
rose has a pretty detailed answer; not alot of interaction allowed in it, but a pretty detailed answer nonetheless :)
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