Question

Obtain all zeros of the polynomial f(x) = 2x⁴ + x³ – 14x² – 19x – 6, if two of its zeros are –2 and –1.

Answer 1

Given: f(x) = 2x⁴ + x³ – 14x² – 19x – 6, with zeros x = –2 and x = –1.

Step-wise calculation:

1. From the given zeros, (x + 2) and (x + 1) are factors.

Their product is x² + 3x + 2.

2. Divide f(x) by x² + 3x + 2 by equating coefficients:

Let quotient = ax² + bx + c.

(x² + 3x + 2)(ax² + bx + c) = ax⁴ + (3a + b)x³ + (2a + 3b + c)x² + (2b + 3c)x + 2c. 

Match with f(x):

a = 2

3a + b = 1

⇒ b = 1 – 3a

= 1 – 6

= –5

2a + 3b + c = –14 

⇒ 4 – 15 + c = –14 

⇒ c = –3

Checks: 2b + 3c = –10 – 9 = –19; 2c = –6 

So quotient = 2x² – 5x – 3.

3. Solve 2x² – 5x – 3 = 0. 

Discriminant Δ = (–5)² – 4 × 2 × (–3) 

= 25 + 24

= 49, `sqrt(Δ)` = 7. 

Roots: `x = (5 ± 7)/(2 xx 2)` 

⇒ `x = 12/4 = 3` or `x = (5 - 7)/4 = -2/4 = -1/2`.

The four zeros of f(x) are x = –2, –1, 3, and `-1/2`.