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okay, I struggle hard core with these and I don't know why. so I need to put 0=6^2p+5p+1 into intercept form
 one year ago
 one year ago
okay, I struggle hard core with these and I don't know why. so I need to put 0=6^2p+5p+1 into intercept form
 one year ago
 one year ago

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TuringTestBest ResponseYou've already chosen the best response.1
slope intercept of\[0=6^2p+5p+1\]are you sure you typed it right?
 one year ago

karinewoods17Best ResponseYou've already chosen the best response.0
try changing all the p's into x's if you see it better that way.
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
oh no! i meant for the p to be squared!
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
it should be 6p^2+5p+1=0 and i need to change that into xintercept form
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
find the xintercepts you mean then, so solve the equation
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
I'm not allowed to use the quadratic formula. i can only factor it
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
6p^2+5p+1=0 ( _p + _ )( _p + _ )=0 we need to fill in the blanks such that the last terms multiply to 1 and the first terms multiply to 6
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
okayyy and how do i find those solutions mathmatically?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
what are the only two integers that can multiply to be 1 ?
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
1 aaand 1 of course
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
good observation that it could be negative, but since everything is postive here we will just assume it is positive for now (_p+1)(_p+1) what pairs of integers multiply to 6 ?
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
1 & 6 2&3 aanndd thats it
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
right, and since we know that (3p+1)(2p+1)=0 that implies that either 3p+1=0 or 2p+1=0, or both so solve those two equations and you get your two values of x for which y is zero, i.e. where the graph intercepts the xaxis.
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
so (3p+1)(2p+1)=0 is tha answer? and i can use this process for every problem?
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you should always doublecheck that you get the correct cross term to be sure you have done it right (3p+1)(2p+1)=0 6p^2+3p+2p+1=0 6p^2+5p+1=0 ^^ cross term when factoring something that looks like ax^2+bx+c=0 sometimes you can find numbers that multiply to a and c, but do not produce the correct b, so you should FOIL out what you have to be sure also, you are not done with the statement (3p+1)(2p+1)=0 you still need to follow what I said above and solve for p: since we know that (3p+1)(2p+1)=0 that implies that either 3p+1=0, or 2p+1=0, or both equal zero so solve those two equations and you get your two values of x for which y is zero, i.e. where the graph intercepts the xaxis.
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
dude, how I wish you were my math teacher
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
hehe, I do too sometimes welcome!
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
oh, typo... should say: so solve those two equations and you get your two values of p for which the function is zero, i.e. where the graph intercepts the paxis.
 one year ago

christinaxxxBest ResponseYou've already chosen the best response.0
haha, yes i understand! thank you so much!
 one year ago
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