Find the zeros of the polynomial f(x)=2x3-7x2-3x+18
try factors of 18
f(1) = 2-7-3+18 ≠ 0
f(-1) ≠ 0
f(2) = 16 - 28 - 6 + 18 = 0 , so x-2 is a factor, and x = 2 is a zero
use either long algebraic division or synthetic division to divide f(x) by x-2 to get a quadratic. Solve this quadratic to find the other zeros.
To find the zeros of a polynomial, we need to solve the equation f(x) = 0. In this case, our polynomial is f(x) = 2x^3 - 7x^2 - 3x + 18.
Step 1: Start with the Rational Root Theorem.
The Rational Root Theorem states that if a polynomial has a rational root, it can be written in the form p/q, where p is a factor of the constant term (in this case, 18) and q is a factor of the leading coefficient (in this case, 2). So, the possible rational roots for f(x) are ±1, ±2, ±3, ±6, ±9, ±18.
Step 2: Test the potential roots.
Using synthetic division or long division, we can evaluate if our potential roots are indeed zeros of the polynomial. Let's start by testing x = 1.
1 │ 2 -7 -3 18
│ 2 -5 -8
└─────────────
2 -5 -8 10
Since the remainder is not equal to zero, x = 1 is not a root of the polynomial.
Let's continue testing the other potential roots.
-1 │ 2 -7 -3 18
│ -2 9 -6
└─────────────
2 -9 6 12
-2 │ 2 -7 -3 18
│ -4 22 -38
└─────────────
2 -11 19 -20
-3 │ 2 -7 -3 18
│ -6 39 -108
└─────────────
2 -13 36 -90
-6 │ 2 -7 -3 18
│ -12 114 -624
└─────────────
2 -19 111 -606
-9 │ 2 -7 -3 18
│ -18 135-1002
└─────────────
2 -25 132 -984
-18│ 2 -7 -3 18
│ -36 630+684
└─────────────
2 -43 627 702
We can see that none of the potential roots are zeros of the polynomial, as the remainder is non-zero for all of them.
Therefore, the polynomial f(x) = 2x^3 - 7x^2 - 3x + 18 does not have any rational zeros.