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

User Goobering
by
6.9k points

2 Answers

3 votes

Answer:


  • 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.
User Mpeac
by
7.5k points
2 votes

Answer:

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

User Manuel Pintaldi
by
7.0k points