Question

PLEASE HELP ! Find the zeros of the given function.

f(x) = x*- 6x3 + 2x2 + 26x + 17
***Look at the picture***

Answer 1

x = -1, 4 + i, 4 - i

Explanation:

Possible rational roots are:

+/- 1, +/- 17

Using trial method:

f(-1) = (-1)⁴ - 6(-1)³ + 2(-1)² + 26(-1) + 17

f(-1) = 0

f(1) = (1)⁴ - 6(1)³ + 2(1)² + 26(1) + 17

f(1) = 40

f(-17) = (-17)⁴ - 6(-17)³ + 2(-17)² + 26(-17) + 17

f(-17) = 113152

f(17) = (17)⁴ - 6(17)³ + 2(17)² + 26(17) + 17

f(17) = 55080

This implies that x = -1 is a root, with multiplicity of 2 or 4

Let the other quadratic factor be g(x)

g(x) = [x⁴ - 6x³ + 2x² + 26x + 17] ÷ (x + 1)²

g(x) = [x⁴ - 6x³ + 2x² + 26x + 17] ÷ (x² + 2x + 1)

g(x) = [x²(x² + 2x + 1) - 8x(x² + 2x + 1) + 17(x² + 2x + 1)] ÷ (x² + 2x + 1)

g(x) = x² - 8x + 17

g(x) = 0

x = [-(-8) +/- sqrt[(-8)² - 4(1)(17)]/2

x = [8 +/- 2i]/2

x = 4 +/- i

Answer 2

-1, 4+i, 4-i

Explanation:

x^4- 6x^3 + 2x^2 + 26x + 17

Using the rational root theorem

we see if 1, -1, -17 or 17 are roots

Check and see if 1 is a root

1^4- 6(1^3) + 2(1^2) + 26(1) + 17=0

1-6+2+26+17 does not equal 0 1 is not a root

-1

1^4- 6(-1^3) + 2(1^2) + 26(-1) + 17=0

1 +6 +2 -26+17 = 0

-1 is a root

Factor out (x+1)

(x+1) ( x^3-7x^2+9x+17)

Using the rational root theorem again on x^3-7x^2+9x+17

Checking -1

-1 -7 -9 +17=0

-1 is a root

(x+1) (x+1) (x^2-8x+17)

Using the quadratic on the last

8 ±sqrt(8^2 - 4(1)17)

--------------------------------

2

gives imaginary roots

4±i