Solve this polynomial for all the real zero's!!?

Question

X^5-25x^3+144x=0 Help me solve this polynomial for all the real zero's!!?

Answer 1

Given polynomial is

Divide each side by x.

Use factor method

Substitute t value

Take square root on each side

Therefore x = 3 , -3 , 4 and -4.

 

Comments

x = 0 is one more factor for the above expression.
x=0,3,−3,4,−4

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.

The polynomial is P (x ) = x⁵ - 25x³+ 144x = 0.

If p/q is a rational zero, then p  is a factor of 0 and 144 and q  is a factor of 1.

The possible values of p  are   0.

The possible values for q  are ± 1.

So, p/q = 0..

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

p/q	1	0	-25	0	144	0
0	1	0	- 25	0	144	0

Since f (0) = 0,  x  = 0is a zero. The depressed polynomial is  x⁴- 25x² + 144.

 

If p/q is a rational zero, then p  is a factor of 0 and 144 and q  is a factor of 1.

The possible values of p  are  ± 1, ± 2, ± 3, and ± 4.

The possible values for q  are ± 1.

So, p/q = ± 1, ± 2, ± 3, and ± 4.

Answer 3

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

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

p/q	1	0	-25	0	144
- 1	1	- 1	- 24	24	120
1	1	1	- 24	- 24	120
- 3	1	- 3	- 16	48	0

Since f (- 3) = 0,  x  = - 3 is a zero. The depressed polynomial is  x³- 3x² - 16x + 48.

 

If p/q is a rational zero, then p  is a factor of 48 and q  is a factor of 1.

The possible values of p  are  ± 1, ± 2, ± 3, and ± 4.

p/q	1	- 3	-16	48
3	1	0	- 16	0

 

Answer 4

Contd.....

Since f (3) = 0,  x  = 3 is a zero. The depressed polynomial is  x²- 16.

By factor by grouping

(x - 4)(x + 4) = 0

Apply zero product property.

x - 4 = 0 and x + 4 = 0

x = 4 and x = - 4.

Rational zeros of x⁵ - 25x³+ 144x = 0 are at x = 0 , - 3, 3, - 4 and 4.