3x^4-19x^3+39x^2-29x+6 what is the 3 rational roots

Question

i need the answer.

Answer 1

Given polynomial is 3x^4-19x^3+39x^2-29x+6

By synthatic division,

Answer 2

Identify Rational Zeros 

3x ⁴ -19x ³ + 39x ² – 29x + 6 = 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 6 and q  is a factor of 3.

The possible values of p  are   ± 1,  ± 2, ± 3,± 6 .

The possible values for q  are ± 1, ± 3.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2, ± 3, ± 1/3, ± 2/3.

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

p/q	3	-19	39	-29	6
-1	3	-22	61	90	-84
1	3	-16	23	-6	0

Since f (1) = 0, x = 1 is a zero. The depressed polynomial is  3x ³ -16x ² + 23x - 6 = 0.

Answer 3

Contonuous...

Now 3x ³ -16x ² + 23x - 6 = 0.

If p/q is a rational zero, then p is a factor of 6 and q is a factor of 3.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1, ± 2, ± 3, ± 1/3, ± 2/3 .

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

p/q	3	-16	23	-6
1	3	-13	10	4
-1	3	-19	42	-48
-2	3	-22	67	-140
2	3	-10	3	0

Since f (2)  =  0,  x = 2 is a zero. The depressed polynomial is  3x² – 10x + 3 = 0.

Now factorise 3x² – 10x + 3 = 0.

3x ² – 9x - x + 3 = 0

3x (x - 3) - 1(x - 3) = 0

(x - 3) (3x - 1) = 0

Apply zero product property.

x - 3 = 0 and 3x - 1 = 0

x = 3 and 3x = 1

x = 3 and x = 1/3

Rational roots of the 3x ⁴ -19x ³ + 39x ² – 29x + 6 = 0 are x = 1 , 2 , 3 and 1/3.