Solving Third-Degree Polynomials, etc.

Question

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

Answer 1

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

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

Comments

You are really good at math!!! :)

Answer 2

 

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  5x² – 2x + 1 = 0

Since the depressed polynomial of this zero, 5x² – 2x + 1, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation.

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.

 

Answer 3

 

44. x⁴ + 4x³ + 2x² – x + 6 = 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 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  x³ + 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  x² – x + 1 = 0

 

Answer 4

 

Contd......

Since the depressed polynomial of this zero, x² – x + 1, is quadratic, use the Quadratic Formula to find the roots of the related quadratic equation x² – x + 1.

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

 

 

Comments

Could you explain the synthetic division in detail for x3 + x2 - x + 15 = 0 for x = -3, please show the work. Thanks in advance.

Answer 5

 

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.

 

Comments

Thank you!!! You are really good at math.... :)