how to solve x^5-8x^3-9x=o by using the theorem of algebra

Question

i need to solve x^5-8x^3-9x=0

Answer 1

Therom of algebra(factor therom)

Let f(x) be a polynomial such that f(c) = 0 ,then x-c is factor.

Given polynomial x^5-8x^3-9x = 0

x(x^4-8x^2-9) = 0

x(x^4-9x^2+x^2-9) = 0

x(x^2(x^2-9)+1(x^2-9)) = 0

x(x^2-9)(x^2+1) = 0

x = 0 ,x^2-9 = 0, x^2+1 = 0

x^2 = 9, x^2 = -1

Squre root negitive is not determined.

So thesolutions are x = 0, x^2 = 9

Solution x = 0,3,-3

 

Answer 2

The polynomial x⁵ - 8x³  - 9x  = 0

x (x⁴ - 8x² - 9) = 0

Apply zero product property

x = 0 and x⁴ - 8x² - 9 = 0

x = 0 is one solution of x⁵ - 8x³  - 9x  = 0

Now solve x⁴ - 8x² - 9 = 0

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.

If p/q is a rational zero, then p is a factor of 9 and q is a factor of 1.

The possible values of p are   ± 1, ± 3,± 9.

The possible values for q are ± 1

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ±3, ± 9

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

p/q	1	0	-8	0	-9
1	1	1	-7	-7	-16
-1	1	-1	-7	7	-16
3	1	3	1	3	0

Since f (3) = 0,   x = 3 is a zero. The depressed polynomial is  x³ + 3x² + x + 3 = 0.

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,   ± 3.

Answer 3

Continuous....

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

p/q	1	3	1	3
1	1	4	5	8
3	1	6	19	60
-1	1	2	-1	4
-3	1	0	1	0

Since f (-3) = 0, x = -3 is a zero. The depressed polynomial is  x² + 1 = 0

x² = - 1

Apply squre root on each side.

x = ±√-1

x = ±√(i)² 

x = ± i

The polynomial x⁵ - 8x³  - 9x  = 0 have three real roots and two imaginary roots.

Solutions x = 0, 3, -3, i, -i.