Solve each quadratic equation using factoring?? Pleasee help :) 2x^2-x-21=0
I thought you were gonna do your own work now...
i changed the question
look at -45 think of multiples sets of -45 that when added is equal to 4 multiple sets -1 45 -3 15 -5 9 etc etc etc
@lisapeajac after this question is the last two problems that i'll do mysself
No more help from me. Good night.
but i was gon show you that i knew how to do it @lisapeajac
@Tutu Please show us what you've done for the question so far so that we know where you're stuck at. Thanks!
2x^2-x-21=0 (2x-3) (x-7)=0 2x-3=0 x-7=0 x=3/2 or x=7 Is that right????
Nope, not really. When you expand (2x-3) (x-7), it doesn't give you 2x^2-x-21.
oh well there you go i'm stuck on that part
2x^2-x-21=0 just looking at the signs + - - since the last term is negative, that means the middle term must be a combination of a positive and negative value because the only way to get a negative number by multiplying is by multiplying a negative number to a positive number
for quadratic equations Ax^2+Bx+C=0 we look at only the A and C terms and start writing down their multiple pairs A= 2 multiple pairs of 2 2 1 C=-21 multiple piars of -21 1 -21 3 -7 7 -3 21 -1 now we determine what 2 multiple pairs ( one from each) when multiplied together and added will give you B, in this case -1 now since its a negative number, we know that the larger number must be negative example, say i take the multiple pair (2 1) and (7 -3) combining it would consist of 2*7 + 1*-3 = 11 or 2(-3)+1*7= =1 since none of those combination gave us 1, they are not the correct factors, but note that we got 1, which is just the negative of the answer we're looking for this means that we got the signs wrong so the multiple pair would be (2 1) and (-7 3) now that we have the multiple pair we write out the factorization (2x-7)(x+3) notice where the numbers are placed the -7 is placed with the 2 mainly because you dont want to multiply those 2 numbers together
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