What are the possible rational zeros of f(x) = x4 + 2x3 – 3x2 – 4x + 18 ?
The number of roots of any polynomial is given as the order of the polynomial. For n-order polynomial, the minimum number of roots are "n'. Given, f(x) = x^4 + 2x^3 – 3x^2 – 4x + 18, it has minimu of four roots. The given expression has the following roots ± 1, ± 2, ± 3, ± 6, ± 9, ± 18
Answer:
The possible rational roots are ±1, ±2, ±3, ±6, ±9, ±18
Step-by-step explanation:
Given: \(f(x)=x^4+2x^3-3x^2-4x+18\)
We need to find all possible rational root of the given function.
Using rational root theorem test.
We have to factor the leading coefficient and constant term
Leading coefficient : 1
Constant term : 18
Factor of 18 (p): 1,2,3,6,9,18
Factor of 1 (q) : 1
We will divide each factor of 18 by each factor of 1.
Possible rational root: \(\pm\dfrac{p}{q}\)
Rational roots are:
\(\Rightarrow \pm1,\pm2,\pm3,\pm6,\pm9,\pm18\)
Hence, The possible rational roots are ±1, ±2, ±3, ±6, ±9, ±18