What are the possible rational zeros of f(x) = x4 + 6x3 − 3x2 + 17x − 15? ± 1, ± 3, ± 5, ± 15 ± 1, ± 1 over 3, ± 1 over 5, ± 1 over 15 ± 1, ± 3, ± 6, ± 15, ± 17 ± 1, ± 1 over 3, ± 1 over 6, ± 1 over 15, ± 1 over 17
@Nnesha
so did you graph it?
`Rational Roots Test` we have to find possible zeros by testing each number from p/q fraction \[\rm \frac{ p }{ q }=\frac{ \pm factors~of~constant~term }{\pm factors~of~leading ~coefficient }\]
so what's the leading coefficient and constant term in that equation ? or as mentioned above you can just graph it zeros (x-intercept ,solution) are points where line intersect the x-axis
um... i have no idea ;/ I'm so lost in this.
how do i find the factors
well i just noticed the graph will give you EXACT solution but we need POSSIBLE
so go with p/q method what is the leading coefficient in that equation ? leading coefficient = coefficient of highest degree variable
and what is the constant term ?
x^4
has the highest degree..
that's same as 1x^4 so one is leading coefficient
ok
and constant term ?
what is that ?
constant term ( number without any variable )
15
or would it be -15 ?
@Nnesha
sorry my computer was shutdown and yes that's correct
what are the factors of 15 ?
1,3,5,15
right \[\rm \frac{ p }{ q }=\frac{ \pm1, \pm 3, \pm 3 , \pm5 , \pm 15 }{\pm 1 }\] now remember each number at the numerator is dividing by 1 simplify each fraction to eliminate the identical values
that's it!
so its A ?
that's correct
Thank you
yw
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