x^4-x^3-6x^2+4x+8 what kind of substitution was used to get the real zeros

Question

I came up with the real zeros are -2,-1 and 2  and put the use of synthetic division, but I can't find how I did this. - any ideas?

Answer 1

The function x⁴ - x³ - 6x² + 4x + 8 = 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 8 and q is a factor of 1.

The possible values of p  are   ± 1,   ± 2, ± 4 and ± 8.

The possible values for q  are ± 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,   ± 2, ± 4 and  ± 8.

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

p/q	1	-1	-6	4	8
1	1	0	-6	-2	6
-1	1	-2	-4	8	0

Since f (-1)  =  0,   x  = - 1 is a zero. The depressed polynomial is  x³ - 2x² – 4x + 8 = 0.

If p/q is a rational zero, then p is a factor of 8 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,  ± 2, ± 4 and  ± 8.

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

p/q	1	-2	-4	8
1	1	-1	-5	3
-1	1	-3	-1	9
-2	1	-4	4	0

Since f (-2)   =   0,   x = -2 is a zero. The depressed polynomial is  x² – 4x + 4 = 0

Since the depressed polynomial of this zero, x² – 4x + 4, is quadratic, solve the roots of the related quadratic equation

x² – 4x + 4 = 0

(x - 2)²  = 0

(x - 2)( x - 2) = 0

x = 2 and x = 2

Real zeros of x⁴ - x³ - 6x² + 4x + 8 = 0 are at x = -1 , - 2 ,2 and 2.