What is a quartic function with only the two real zeros given? x = -4 and x = -1 A. y = x^4 + 5x^3 + 5x^2 + 5x + 4 B. y = x^4 - 5x^3 - 5x^2 - 5x - 4 C. y = -x^4 + 5x^3 + 5x^2 + 5x + 4 D. y = x^4 + 5x^3 + 5x^2 + 5x - 5
just substitute your both x and find which one gives 0 for both
I dont get it. Im just confusing myself.
but how will you know the OTHER two zeros are REAL or complex?
because only for one equation these two are zeros,for other 3 even these are not zeros
@hartnn it says with ONLY the two. Some of these can have more than 2 real roots. The systematic way i'd go around doing this would to 1) first substitute x =-4 and x=-1into the equations to see which ones actually have roots of -4 and -1(aka y=0)
i'm with @dpalnc . You should carry out a systematic way to solve these
i did that only and found out that only one equation has y=0......
so you did do a systematic way....:)
what i did was testing of options :P
lets ask whether asker understood....Emily?
nothing wrong with doing it by elimination... just becareful...:)
hartnn, all i'm saying is what would you have done if there was two with roots -4 and -1
will try to factorize...
more specifically,first i will divide that by x+1 and x+4....because they are factors....and get the quadratic expression,which would be easier to solve for....
could use descartes rule also =]
i don't know what that is..
i didn't know it either until a few weeks ago lol
doesn't give exact no of roots...
So.. Whats the answer?
its A :P
Thanks!
yes but it tells you the possible amount of negative roots .. if you get 2 posible negative roots. and you have 2 negative roots it means that there is only 2 non complex roots. if you look there is 0 possible roots due to no switching in signs
0 possible positive roots
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