Finding all real zeros of

Question

x^4 - x^3 - 6x^2 + 4x + 8?

 

I'm stuck on this part:

(x-2)(x+1)(1x^2 + 0x - 4)

^I got this as my final solution and was wondering if the 1x^2 + 0x - 4 part is right? So confused!!!

So -1 works so I used synthetic division, got 1x^3 - 2x^2 - 4x + 8. So then I plugged in 2 to see if it worked and it equals 0, so 2 works. So I use synthetic division again and divide 1x^3 - 2x^2 - 4x + 8 by 2 and I got 1x^2 + 0x - 4 ...

so i put everything together, which is (x-2)(x+1)(1x^2 + 0x - 4) but I feel I am doing something wrong. what the hell am I doing wrong??

 

Answer 1

The polynomial 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	0	-4	0

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

Solve the equation  x² –  4 = 0

x²  = 4

Aplly squre root on each side.

x = ± 2.

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