Please help I don't understand this at all: Approximate the real number solution(s) to the polynomial function f(x) = x3 + 4x2 + x − 6?
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Hi @Nnesha, do you think you can help?
We don't have a perticular formula for checking solution of a cubic polynomial. So we need to check it... Start with smallest factor of 6 ...X=1 ,if you get f(1) =0 ,than 1 is its solution
looks like p/q method would work
i think i have to find the factors of -6 and put them over the factors of 1 right?
\[\rm \frac{ p }{ q }=\frac{ \pm factors~ of~constant }{ \pm factors~of ~leading~coefficient }\]
Yes...
yep that's correct and then test each possible solution by either using synthetic division or replacing x for its value
so i would end up with \[\frac{ 3 }{ 1 }, \frac{ 2 }{ 1 }, \frac{ 6 }{ 1 }, \frac{ 1 }{ 1 }\]
it should be plus/minus
of course sorry i forgot
then i would plug either +/-3, +/-2, +/- 6, +/-1 into my function and observe which number results in f(x) = 0. Right?
yep and remember if there are duplicates numbers after dividing p/q u should eliminate it
Here, f(1)=0 Means (x-1) is its factor. You can find remaining factor by dividing f(x) by (x-1). You will get a quadratic equation And you know how to solve them.
so would i synthetically divide x-1 by x^3 + 4x^2 + x - 6 (I'm terrible at math so i apologize)
Yes,
or you can find \[\rm f(\pm1),f(\pm2) f(\pm 3),f(\pm6)\] and see which one give you 0 as final answer that would be the real zero ~zeros (x-intercept) is a point where the graph intersect the x-axis when y=0 and in this question f(x) is same as y
Ill try both your methods although the second seems more time consuming so ill try synthetic division first
true that if you are familiar with the synthetic division method then ofc do that first.
ok so i got x-1, x+3, and x + 2
i synthetically divided the inverse of each number and they all gave me a reminder of 0
looks good to me.
Thank you so much for the help, you both were wonderful help! :) XD
np. good work btw keep it up!
Ill go ahead and close the question now.
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