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Finding all real zeros of

0 votes

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??

 

asked Jul 23, 2014 in ALGEBRA 2 by anonymous

1 Answer

0 votes

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 anxn + an  1xn – 1 + ... + a1x + a0 = 0, then p is a factor of a0 and q is a factor if an.

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  x–  4 = 0

Solve the equation  x–  4 = 0

x = 4

Aplly squre root on each side.

x = ± 2.

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

answered Jul 23, 2014 by david Expert

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