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 = [tex]\frac{1}{2}[/tex] ← with multiplicity 2
x - 2 = 0 ⇒ x = 2 ← with multiplicity 2
Hence the roots are
{ 4, [tex]\frac{1}{2}[/tex], 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 ± [tex]\sqrt{2^2-(4(1)(3)}[/tex] ) / 2
= ( - 2 ± [tex]\sqrt{4-12}[/tex] ) / 2
= ( - 2 ± [tex]\sqrt{-8}[/tex] ) / 2
= (- 2 ± 2i[tex]\sqrt{2}[/tex] ) / 2
= - 1 ± i[tex]\sqrt{2}[/tex]
Hence roots are
{ - 2, - 1 + i[tex]\sqrt{2}[/tex], - 1 - i[tex]\sqrt{2}[/tex] }
