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Solving Third-Degree Polynomials, etc.

+6 votes
Solve each polynomial in the complex number system.

How would you do so when grouping isn\'t an option?

 

42. x^3+x^2-x+15=0

43. 5x^3-12x^2+5x-2=0

44. x^4+4x^3+2x^2-x+6=0
asked Dec 25, 2012 in ALGEBRA 2 by alg2trig Rookie

5 Answers

+7 votes

Identify Rational Zeros  

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

42. x3+x2-x+15 = 0

If p/q is a rational zero, then p is a factor of 15 and q is a factor of 2.

The possible values of p are   ± 1,   ± 3, and   ± 5.

The possible values for q are ± 1.

So, p/q = ± 1,   ± 3, ± 5.

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

p/q 1 1 -1 15
1 1 2 1 16
3 1 4 11 48
5 1 6 29 160
-1 1 0 -1 16
-3 1 -2 5 0

Since f(-3)   =   0,   x   =   -3 is a zero. The depressed polynomial is   x2-2x+5 = 0

Since the depressed polynomial of this zero, x2-2x+5, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation

depressed polynomial x^2-2x+5 = 0 roots using quadratic formula

The function has one real zero at x = -3 and two imaginary zeros at x = 1 + 2i and x = 1 - 2i.

answered Dec 28, 2012 by steve Scholar
You are really good at math!!! :)
+3 votes

 

45. 5x3-12x2+5x-2 = 0

 
If p/q is a rational zero, then p is a factor of 2 and q is a factor of 5.

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

The possible values for q are ± 1 and   ± 2.

So, p/q = ± 1,   ± 2,   ±1/5,   ± 2/5

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

p/q

5

-12

5

-2

1

5

-7

-2

-4

2

5

-2

1

0

Since f(2)   =   0,   x  =  2 is a zero. The depressed polynomial is  5x– 2x + 1 = 0

Since the depressed polynomial of this zero, 5x– 2x + 1, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation.

depressed polynomial  5x2 – 2x + 1 = 0 roots

The function has one real zero at x = 2 and two imaginary zeros at x = 1/5 + i/5 and x = 1/5 - i/5.

 

answered Dec 28, 2012 by steve Scholar
+4 votes

 

44. x4 + 4x+ 2x– x + 6 = 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 6 and q is a factor of 1.

The possible values of p are   ± 1,   ± 2, ± 3.

The possible values for q are ± 1

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,   ± 2,   ±3

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

p/q

1

4

2

-1

6

1

1

5

7

6

12

2

1

6

14

27

60

3

1

7

23

68

210

-1

1

3

-1

0

6

-2

1

2

-2

3

0

Since f(-2)   =   0,   x   =   –2 is a zero. The depressed polynomial is  x+ 2x2 – 2x + 3 = 0.

 

If p/q is a rational zero, then p is a factor of 3 and q is a factor of 1.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,   ± 3.

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

p/q

1

2

-2

3

1

1

3

1

4

3

1

5

13

42

-1

1

1

-3

6

-3

1

-1

1

0

Since f(-3)   =   0,   x = -3 is a zero. The depressed polynomial is  x– x + 1 = 0

 

answered Dec 28, 2012 by steve Scholar
+5 votes

 

Contd......

Since the depressed polynomial of this zero, x– x + 1, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation x– x + 1.

depressed polynomial  x2 – x + 1 = 0 roots using quadratic formula

The function has one real zero at x = -2 and x = -3, two imaginary zeros at x = 1/2 + √3i/2 and 

x = 1/2 - √3i/2

 

 

answered Dec 28, 2012 by steve Scholar
Could you explain the synthetic division in detail for x3 + x2 - x + 15 = 0 for x = -3, please show the work. Thanks in advance.
+5 votes

 

The polynomial is x3 + x2 - x + 15 = 0

Start out with the synthetic division algorithm. Start as usual by bringing down the 1:

-3  |     1             1         -1           15

     |_________________________                                                           


              1

 Multiply the 1 by -3 and put it diagonally under the next coefficient.

Add -3 and 1, getting -2 and write this on the bottom of the line.

 

-3  |     1             1         -1           15

     |                     -3                                   

            1             -2                 

Multiply the -2 by -3 and put it diagonally under the next coefficient

Add -1 and 6, getting 5 and write this on the bottom of the line

 

-3  |     1             1         -1           15

      |                  -3             6                       

            1            -2            5

 

Multiply the -3 by 5 and put it diagonally under the next coefficient. Add 15 and -15, getting 0 as remainder and write this on the bottom of the line.

 

 

-3  |     1             1         -1           15

    |                   -3           6           -15    

            1           -2         5               0

Since the remainder is zero, then    x = -3 is indeed a zero of x3 + x2   – x + 15 = 0.

 

answered Dec 28, 2012 by steve Scholar
Thank you!!! You are really good at math.... :)

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