Answer:
[tex]\large \boxed{\sf \ \ \ -1, \ -5, \ -\dfrac{1}{4} \ \ \ }[/tex]
Step-by-step explanation:
Hello,
Let's determine the possible rational zeros of this polynomial function using the rational zeros theorem:
[tex]P(x) = 4x^4 + 13x^3-49x^2-73x-15[/tex]
First of all, what is the rational zeroes theorem?
If P(x) is a polynomial with integer coefficients
and if (p and q being integer)
[tex]\dfrac{p}{q}[/tex]
is a zero of P(x), meaning
[tex]P(\dfrac{p}{q})=0[/tex]
then p is a factor of the constant term of P(x) and
q is a factor of the leading coefficient of P(x).
How to apply it here?
The constant term of P(x) is -15
The leading coefficient of P(x) is 4
so p is a factor of -15
q is a factor of 4
15 = 1 * 5 * 3
4 = 2 * 2 * 1
q can be 1, 2, 4
-p can be 1, 3, 5, 15
so it gives the following potential solutions
-1, -3, -5, -15
[tex]\dfrac{-1}{2}, \dfrac{-3}{2}, \dfrac{-5}{2}, \dfrac{-15}{2}[/tex]
[tex]\dfrac{-1}{4}, \dfrac{-3}{4}, \dfrac{-5}{4}, \dfrac{-15}{4}[/tex]
Let's compute P(x) for x in this list of potential solutions
x P(x)
-1 0
-3 -264
-5 0
-15 148680
-0.5 7.875
-1.5 -39.375
-2.5 -185.625
-7.5 4948.125
-0.25 0
-0.75 7.96875
-1.25 -15.9375
-3.75 -324.84375
It gives -1, -5 and -0.25
Conclusion
The possible rational zeroes of P(x) are
-1
-5
[tex]\dfrac{-1}{4}[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
