Solving Third-Degree Polynomials, etc.

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

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.

You are really good at math!!! :)

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.

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.

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

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.

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

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

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.