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How do you solve equations over a set of complex numbers?

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2x^4 - 3x^3 - 24x^2 + 13x +12? 

asked Nov 18, 2014 in PRECALCULUS by anonymous

1 Answer

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The equation 2x4- 3x³-24x2 + 13x+12 = 0

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation anxn + an  1 xn–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 is a factor of 2 and q  is a factor of 12.

The possible values of p  are   ± 1 and ± 2.

The possible values for q are ± 1, ±2, ±3, ±4, ±6 and ±12.

Possible p/q values are ± 1, ± 1/2, ± 1/3, ± 1/4, ± 1/6, ± 1/12, ± 2, ± 2/3, ± 2/6, ± 2/12

Let us consider a test point p/q = ± 1

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

p/q

2

-3

-24

13 12

1

2

-1

-25

-12

0

-1

2

-5

-19

32

-20

Since f (1) = 0,  x = 1 is a zero. The reduced polynomial is  2x3 - x 2 - 25x - 12= 0.

Consider the polynomial x3 - x 2 - 25x - 12= 0.

If p/q is a rational zero, then is a factor of 1 and q  is a factor of 12.

The possible values of p  are   ± 1.

The possible values for q are ± 1, ±2, ±3, ±4, ±6 and ±12.

Possible p/q values are ± 1, ± 1/2, ± 1/3, ± 1/4, ± 1/6 and ± 1/12

Let us consider a test point p/q = ±1 and - 1/2

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

p/q

2

-1 -25 -12

1

2

1 -24 -36

- 1

2

3

-22

-3

- 1/2

2

-2

-24

0

Since f (-1/2) = 0,  x = -1/2 is a zero. The reduced polynomial is  2x 2 - 2x - 24= 0.

2x 2 - 2x - 24= 0.

2(x2 - x - 12)= 0

x2 - 4x + 3x - 12 = 0

x(x - 4) + 3(x - 4) = 0

(x + 3)(x - 4) = 0

x = -3 and 4

Therefore the values of x are 1, -1/2, -3 and 4.

answered Nov 18, 2014 by Lucy Mentor

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