Question

How do I find the zeros of x^4-10x^3-66x^2-94x-39

Answer 1

Answer: x = -1, -3, 7 + √62, 7 - √62

Step-by-step explanation:

q                                    p

x⁴ - 10x³ - 66x² - 94x - 39

$\frac{p}{q}$ = +/- $\frac{1*3*13*39}{1}$

possible rational factors: 1, -1, 3, -3, 13, -13, 39, -39

Use synthetic division or long division to see which factor will leave a remainder of 0.

try x + 1 = 0  ⇒   x = -1

-1   |  1     -10     -66     -94     -39

    |  ↓     -1        11        55      39

       1     -11      -55     -39       0

(x + 1)(x³ - 11x² - 55x - 39)        

next, try x + 3 = 0  ⇒   x = -3   for the new polynomial

-3   |  1     -11     -55     -39

     |  ↓     -3       42     39

        1     -14      -13       0  

(x + 1)(x + 3)(x² - 14x - 13)

Lastly: find the zeros by setting each factor equal to zero and solve.

x + 1 = 0  ⇒  x = -1

x + 3 = 0   ⇒ x = -3

x² - 14x - 13 = 0

$x = \frac{-(b) +/- \sqrt{(b)^{2} -4(a)(c)}}{2(a)}$

  $= \frac{-(-14) +/- \sqrt{(-14)^{2} -4(1)(-13)}}{2(1)}$

  $= \frac{14 +/- \sqrt{196+52}}{2}$

  $= \frac{14 +/- \sqrt{248}}{2}$

  $= \frac{14 +/- 2\sqrt{62}}{2}$

  $= 7 +/- \sqrt{62}$