Question

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

Answer 1

Answer:

• $x_1 = -2$.
• $x_2 = 6$.

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

Step-by-step 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:

$\displaystyle (x - 6) \cdot (x + 2) = 0\times\frac{1}{4} = 0$.

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

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

Answer:

X=6 , X = -2

Step-by-step 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