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anonymous
 one year ago
8x2 + 64x = 0
anonymous
 one year ago
8x2 + 64x = 0

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e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4I am guessling that is 8x^2... So, what problem are you having with it?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answers choices are x=0 ,x=8 x=0,x=8 x=8,x=0,x=8 x=8,x=64

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4Yes, and what is giving your trouble? This is a tutoring site...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand anything at all im lost can you explain ?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4When you have something = 0 equations, you need to factor it and then find when the factors would be 0. This is sometimes called the roots, zeros, or x intercepts of a polynomial.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4So, if you factor \(8x^2 + 64x\), what do you get?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4Um, that is the first term, but it is not the factors.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok , so can you break this down in help me step by step ?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4Well, do you know what factoring is?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4OK, if I say \(a\times b\times c\) then a, b and c are factors. They are multiplied together. Another common use is prime factorization. Like how 15 has the prime factors of 3 and 5. So that polynomeal is the result of multiplying together something. When you unmultiply it, you are factoring it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x^2 would be the factors ?

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4No, it would be something like \((x\pm h)(x\pm k)\) where h and k are some constants.

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4If I multiply out \((x+1)(x3)\) I get: \(x^2+x3x3\) or: \(x^22x3\) So \((x+1)\) and \((x3)\) are factors of \(x^22x3\). That is why factoring can be called unmultiplying.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ion know i just guessed

e.mccormick
 one year ago
Best ResponseYou've already chosen the best response.4One big clue with this one is that each term has x in it. If you factor out the x, what would you get?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0when someone is helping you, and they ask you a question and you guess, it makes them not want to hep you.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0solve your equation stepbystep. 8x2+64x=0 Factor left side of equation. 8x(x+8)=0 Set factors equal to 0. 8x=0 or x+8=0 This should be more than enough clues :)
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