The document discusses factorizing a polynomial P(x) = 4x^4 + 8x^3 + 5x^2 + x - 3 over the complex field. It is given that P(x) has two rational zeros. The zeros are found to be 1/2 and -3/2. P(x) is then factorized as (2x - 1)(2x + 3)(x^2 + x + 1) = (2x - 1)(2x + 3)(x + (1/2))^2 - (1/4).

1. Factorising Into
Complex Factors
e.g. Given that P x   4 x 4  8 x 3  5 x 2  x  3 has two rational zeros,
find these zeros and factorise P(x) over the complex field.

2. Factorising Into
Complex Factors
e.g. Given that P x   4 x 4  8 x 3  5 x 2  x  3 has two rational zeros,
find these zeros and factorise P(x) over the complex field.
 1  1  1  1 1
P   4   8   5    3
 2   16   8   4  2
0

3. Factorising Into
Complex Factors
e.g. Given that P x   4 x 4  8 x 3  5 x 2  x  3 has two rational zeros,
find these zeros and factorise P(x) over the complex field.
 1  1  1  1 1
P   4   8   5    3
 2   16   8   4  2
0  2 x  1 is a factor

4. Factorising Into
Complex Factors
e.g. Given that P x   4 x 4  8 x 3  5 x 2  x  3 has two rational zeros,
find these zeros and factorise P(x) over the complex field.
 1  1  1  1 1
P   4   8   5    3
 2   16   8   4  2
0  2 x  1 is a factor
P x   2 x  12 x 3  3

5. Factorising Into
Complex Factors
e.g. Given that P x   4 x 4  8 x 3  5 x 2  x  3 has two rational zeros,
find these zeros and factorise P(x) over the complex field.
 1  1  1  1 1
P   4   8   5    3
 2   16   8   4  2
0  2 x  1 is a factor
P x   2 x  12 x 3  5 x  3

6. Factorising Into
Complex Factors
e.g. Given that P x   4 x 4  8 x 3  5 x 2  x  3 has two rational zeros,
find these zeros and factorise P(x) over the complex field.
 1  1  1  1 1
P   4   8   5    3
 2   16   8   4  2
0  2 x  1 is a factor
P x   2 x  12 x 3 5x 2  5 x  3

7.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0

8.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor

9.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3
 rational zeros are and -
2 2

10.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3
 rational zeros are and -
2 2
P x   2 x  12 x  3x 2  1

11.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3 1 3  2 x  6 x
 rational zeros are and -
2 2
1 2 x  1  2 x
P x   2 x  12 x  3x 2  1

12.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3 1 3  2 x  6 x
 rational zeros are and -
2 2
1 2 x  1  2 x
P x   2 x  12 x  3x 2  1 4x
 1 3  ? x  3 x

13.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3 1 3  2 x  6 x
 rational zeros are and -
2 2
1 2 x  1  2 x
P x   2 x  12 x  3x 2  x  1 4x
 1 3  ? x  3 x

14.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3 1 3  2 x  6 x
 rational zeros are and -
2 2
1 2 x  1  2 x
P x   2 x  12 x  3x 2  x  1 4x
 1  3
2
 1 3  ? x  3 x
 2 x  12 x  3 x    
 2  4

15.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3 1 3  2 x  6 x
 rational zeros are and -
2 2
1 2 x  1  2 x
P x   2 x  12 x  3x 2  x  1 4x
 1  3
2
 1 3  ? x  3 x
 2 x  12 x  3 x    
 2  4
 1 3  1 3 
 2 x  12 x  3 x   i  x   i
 2 2  2 2 

16.   3   2  27   5 9   5  3   3
P       
 2  8   4  2
0  2 x  3 is a factor
1 3 1 3  2 x  6 x
 rational zeros are and -
2 2
1 2 x  1  2 x
P x   2 x  12 x  3x 2  x  1 4x
 1  3
2
 1 3  ? x  3 x
 2 x  12 x  3 x    
 2  4
 1 3  1 3 
 2 x  12 x  3 x   i  x   i
 2 2  2 2 
Exercise 5C; 1 to 15 odds, 16 to 20 all