solve by finding all roots: x^4-5x^3+7x^2-5x+6=0

Question

finding all roots.

Answer 1

Given x^4-5x^3+7x^2-5x+6 = 0

By synthatic division

x^4-5x^3+7x^2-5x+6 = (x-2)(x-3)(x^2+1)

Now solve x^2+1

Compare it to quadratic form ax^2+bx+c .

a = 1, b = 0, c = 1.

Roots of x^4-5x^3+7x^2-5x+6 is x = 2,3,i,-i.

Answer 2

x⁴ - 5x³ + 7x² – 5x + 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 1.

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

The possible values for q  are ± 1

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

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

p /q	1	-5	7	-5	6
1	1	-4	3	-1	7
-1	1	6	14	27	60
-2	1	-7	21	-47	100
2	1	-3	1	-3	0

Since f (2) = 0,   x = 2 is a zero. The depressed polynomial is  x³ - 3x² + x - 3 = 0.

Answer 3

Continuous...

Now 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.

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

p /q	1	-3	1	-3
1	1	-2	-1	-4
-1	1	-4	5	-8
3	1	0	1	0

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

Roots of the polynomial  at x = 3 and x = 2 and two imaginary roots at x = i and x = -i.