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

Answers

ANSWER

The correct options are ,

D. -6 with multiplicity of 2

and

F. -2 with multiplicity of 2

EXPLANATION

The multiplicity of the roots of a polynomial function tells us the number of times the roots the roots of the polynomial repeat.

Considering the polynomial function

$f(x) = {(x + 6)}^{2} {(x + 2)}^{2}$

the roots can be found by equating either factors to zero

${(x + 6)}^{2} = 0$

The square tells us that the root
x = - 6

repeats twice. This means it has a multiplicity of 2.

Also

$(x + 2) ^{2} = 0$
tells us that the root
x = - 2

has a multiplicity of 2

Therefore options D and F are the correct options.

adamkinney9345

theres only the two answers 6 with multiplicity 2 and -2 with multiplicity 2

Step-by-step explanation:

jaqsway34

The roots of $f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}$ are $\boxed{6{\text{ with multiplicity 2}}}$ and $\boxed{ - {\text{2 with multiplicity 2}}}$ . Option (d) and option (f) are correct.

Further explanation:

The Fundamental Theorem of Algebra states that the polynomial has n roots if the degree of the polynomial is n.

$f\left( x \right) = a{x^n} + b{x^{n - 1}} + \ldots + cx + d$

The polynomial function has n roots or zeroes.

Given:

The equation is $f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}.$

Explanation:

The given equation is $f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}.$

According to the Fundamental Theorem of Algebra the function has 4 roots.

Solve the above equation to obtain the zeros.

$\begin{aligned}f\left( x \right) &= {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}\\0&= {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}\\0&= {\left( {x - 6} \right)^2}{\text{ or }}{\left( {x + 2} \right)^2} &= 0\\0&= x - 6{\text{ or }}x + 2 &= 0\\6&= x{\text{ or }}x &= - 2 \\\end{aligned}$

The roots are $- 2{\text{ and 6 with multiplicity 2}}$

The roots of $f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}\[\text{are}\: \boxed{6{\text{ with multiplicity 2}}}$ and $\boxed{ - {\text{2 with multiplicity 2}}}$ .Option (d) and option (f) are correct.

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Learn more about inverse of the Learn more about equation of circle Learn more about range and domain of the function

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Polynomial

Keywords: roots, linear equation, quadratic equation, zeros, function, polynomial, solution, cubic function, degree of the function, multiplicity of 1, multiplicity of 2.

KaseyT

D) 6 and 2

F) -2 and 2

lilred58

6 with multiplicity 2

-2 with multiplicity 2

Step-by-step explanation:

We have given that find the roots of:

f (x) = (x- 6)²(x + 2)²

Note that the polynomials are already in the factored form. So we will only make it equal to zero

(x-6)² = 0

The square tells us that the root repeat twice

Move -6 to the R.H.S

Then,

x=6 , 6

(x+2)²=0

Move 2 to the R.H.S

Then,

x= -2, -2

Therefore the correct options are:

6 with multiplicity 2

-2 with multiplicity 2

bugbug26

4 and 6

~6 with multiplicity 2

~ -2 with multiplicity 2

trodmanmalachip2yn26

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

kyrajaudon3183

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

Step-by-step explanation:

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

Equate f(x) and zero 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

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