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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

Mathematics
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slope intercept of\[0=6^2p+5p+1\]are you sure you typed it right?
try changing all the p's into x's if you see it better that way.
oh no! i meant for the p to be squared!

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Other answers:

it should be 6p^2+5p+1=0 and i need to change that into x-intercept form
find the x-intercepts you mean then, so solve the equation
I'm not allowed to use the quadratic formula. i can only factor it
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
okayyy and how do i find those solutions mathmatically?
what are the only two integers that can multiply to be 1 ?
1 aaand 1 of course
or -1 and -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 ?
1 & 6 2&3 aanndd thats it
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 x-axis.
so (3p+1)(2p+1)=0 is tha answer? and i can use this process for every problem?
you should always double-check 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 x-axis.
dude, how I wish you were my math teacher
hehe, I do too sometimes welcome!
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 p-axis.
haha, yes i understand! thank you so much!

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