Question

Please need help on these two cause I’m very lost on them

Answer 1

Answer:

see explanation

Step-by-step explanation:

Given

f(x) = (x - 4)(2x - 1)²(x - 2)²

To find the roots equate f(x) to zero, that is

(x - 4)(2x- 1)²(x - 2)² = 0

Equate each of the factors to zero and solve for x

x - 4 = 0 ⇒ x = 4

2x - 1 = 0 ⇒ x = $\frac{1}{2}$  ← with multiplicity 2

x - 2 = 0 ⇒ x = 2 ← with multiplicity 2

Hence the roots are

{ 4, $\frac{1}{2}$, 2 }

Given

f(x) = x³ + 4x² + 7x + 6

Note that

f(- 2) = (- 2)³ + 4(- 2)² + 7(- 2) + 6 = - 8 + 16 - 14 + 6 = 0

Since f(- 2) = 0 then by the factor theorem x = - 2 is a root and (x + 2) a factor

Using synthetic division

- 2 | 1   4    7    6

         - 2  - 4  - 6

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

     1    2    3    0

Thus

f(x) = (x + 2)(x² + 2x + 3)

Solve x² + 2x + 3 using the quadratic formula

with a = 1, b = 2 and c = 3

x = (- 2 ± $\sqrt{2^2-(4(1)(3)}$ ) / 2

  = ( - 2 ± $\sqrt{4-12}$ ) / 2

  = ( - 2 ± $\sqrt{-8}$ ) / 2

  = (- 2 ± 2i$\sqrt{2}$ ) / 2

  = - 1 ± i$\sqrt{2}$

Hence roots are

{ - 2, - 1 + i$\sqrt{2}$, - 1 - i$\sqrt{2}$ }