Question

Find the root(s) of f (x) = (x- 6)2(x + 2)2.

Answer 1

• 
  `x_1 = -2`.
• 
  `x_2 = 6`.

Assumption:
`f(x)` is defined for all
`x\in \mathbb{R}` (all real values of
`x`.)

Explanation:

Evaluating
`f(x)` for a root of this function shall give zero.

Equate
`f(x)` and zero
`0` to find the root(s) of
`f(x)`.

`f(x) =0`.

`(x - 6) \cdot 2 \cdot (x + 2) \cdot 2 = 0`.

Multiply both sides by 1/4:

`\displaystyle (x - 6) \cdot (x + 2) = 0*(1)/(4) = 0`.

`\displaystyle (x - 6) \cdot (x + 2) = 0`.

This polynomial has two factors:

• (x - 6), and
• (x + 2) = (x - (-2)).

Apply the factor theorem:

• The first root (from the factor (x - 6)) will be
  `x = 6`.
• The second root (from the factor (x - (-2)) will be
  `x = -2`.

Answer 2

X=6 , X = -2

Explanation:

For any polynomial given in factorised form , the roots are determined as explained below:

Suppose we have a polynomial

(x-m)(x-n)(x+p)(x-q)

The roots of above polynomial will be m,n,-p, and q

Also if we have same factors more than once , there will be duplicate roots.

Example

(x-m)^2(x-n)(x+p)^2

For above polynomial

There will be total 5 roots . Out of which 2(m and -p) of them will be repeated. x=m , x=n , x =-p

Hence in our problem

The roots of f(x)= (x-6)^2(x+2)^2 are

x=6 And x=-2