I already know the process of factoring what am having a problem with is the last term.Like in n^2-3n-208 how can I found two factors of 208 that adds up into -3 because 208 is a big number. Is there an easier way to find the factors of 208?. Because I try using the trial and error but it takes too much time and I have a time limit in answering soooo I really need help plzzzzzzzzzzzzz
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You have to calculate discriminant !!!
The discriminant of ax²+bx+c=0 is b²-4ac
if b²-4ac >= 0 we can factor the trinomial.
here with n instead of x for the variable, we have
a = 1
b = -3
c = -208
=> disc = b²-4ac = 9 + 4*208 = 841 = 29² > 0, so the trinomial can be factored
the roots of the quadratic equation are
r1 = (-b - sqrt(disc))/2a = -13
r2 = (-b + sqrt(disc))/2a = 16
then the factorisation is a(x-r1)(x-r2)
here this is : n²-3n+208 = (n+13)(n-16)
People lose a lot of time searching by trial and error when there is this quadratic formula !
Use it for quadratic equations !
The discriminant of ax²+bx+c=0 is b²-4ac
if b²-4ac >= 0 we can factor the trinomial.
here with n instead of x for the variable, we have
a = 1
b = -3
c = -208
=> disc = b²-4ac = 9 + 4*208 = 841 = 29² > 0, so the trinomial can be factored
the roots of the quadratic equation are
r1 = (-b - sqrt(disc))/2a = -13
r2 = (-b + sqrt(disc))/2a = 16
then the factorisation is a(x-r1)(x-r2)
here this is : n²-3n+208 = (n+13)(n-16)
People lose a lot of time searching by trial and error when there is this quadratic formula !
Use it for quadratic equations !