Determine the zeros of

Question

f(x) = x^4 – x^3 + 7x^2 – 9x – 18

Answer 1

f(x) = x⁴ - x³ + 7x² - 9x -18

x⁴ - x³ + 7x² - 9x -18 = 0

Synthetic division theorem, possible roots are ±1, ±2, and ±3.

    -1 | 1    -1    7     -9     -18

        |  0    -1    2     -9      18

        |_______________________

     2 | 1     -2    9    -18     0

        | 0      2    0      18

        |_______________________

         1      0     9      0

By checking , we find that -1 and 2 are a roots.

By the fundamental theorem of algebra.

The remaining root are those (x² + 9) = 0

Subtract 9 from each side.

x² = - 9

Apply square root each side.

x = √(-9) = √(9i²) = ± 3i                   [i² = -1]

So the roots are  -1, 2, 3i and -3i.

Comments

Thank you!!! How would you know the possible roots? any rule to be applied?

Answer 2

Identify Rational Zeros :

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nxⁿ + a_n – 1x^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

The function f (x) = x⁴ - x³ + 7x² - 9x - 18.

If p/q is a rational zero, then p is a factor of 18 and q is a factor of 1.

The possible values of p are   ± 1, ± 2, ± 9.

The possible values for q are ± 1

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2, ± 9.

Make a table for the synthetic division and test possible real zeros.

p/q	1	- 1	7	- 9	18
1	1	0	7	- 2	-16
- 1	1	- 2	9	- 18	0

Since f(- 1) = 0, x = – 1 is a zero. The depressed polynomial is  x³ - 2x² + 9x - 18 = 0.

Answer 3

Contd.............

If p/q is a rational zero, then p is a factor of 3 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2, ± 9.

Make a table for the synthetic division and test possible real zeros.

p/q	1	- 2	9	- 18
1	1	- 1	8	- 10
- 1	1	- 3	12	- 30
2	1	0	9	0

Since f(2) = 0, x = 2 is a zero. The depressed polynomial is  x² + 9 = 0.

The depressed polynomial is  x² + 9 = 0.

 x² = - 9

 x² = (± 3i)²

x = - 3i and x = 3i.

The function has two real zeros at x = -1 and x = 2, two imaginary zeros at x = - 3i and 

x = 3i.