How many real roots does x^4-x^3-9x^2+7x+14=0 have? A) 1 B) 2 C) 3 D) 4

Question

How many real roots does
`x^4-x^3-9x^2+7x+14=0` have?

A) 1
B) 2
C) 3
D) 4

Answer 1

Answer: D) 4

Explanation:

x⁴ - x³ - 9x² + 7x + 14 = 0

possible rational roots are: ±{1, 2, 7, 14}

Try x = -1

-1 | 1 -1 -9 7 14

| ↓ -1 2 7 -14

1 -2 -7 14 0 ← remainder is 0 so x = -1 is a root

Let's try x = 2 with the factored polynomial

2 | 1 -2 -7 14

| ↓ 2 0 -14

1 0 -7 0 ← remainder is 0 so x = 2 is a root

Factored polynomial x² - 7 can be factored to:

(x - √7) (x + √7)

⇒ x = √7 and x = -√7

The roots are x = {-1, 2, √7, -√7) which are all real numbers.