The roots of [tex]f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}[/tex] are [tex]\boxed{6{\text{ with multiplicity 2}}}[/tex] and [tex]\boxed{ - {\text{2 with multiplicity 2}}}[/tex]. 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.
[tex]f\left( x \right) = a{x^n} + b{x^{n - 1}} + \ldots + cx + d[/tex]
The polynomial function has n roots or zeroes.
Given:
The equation is [tex]f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}.[/tex]
Explanation:
The given equation is [tex]f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}.[/tex]
According to the Fundamental Theorem of Algebra the function has 4 roots.
Solve the above equation to obtain the zeros.
[tex]\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}[/tex]
The roots are [tex]- 2{\text{ and 6 with multiplicity 2}}[/tex]
The roots of [tex]f\left( x \right) = {\left( {x - 6} \right)^2} \times {\left( {x + 2} \right)^2}\[\text{are}\: \boxed{6{\text{ with multiplicity 2}}}[/tex] and [tex]\boxed{ - {\text{2 with multiplicity 2}}}[/tex].Option (d) and option (f) are correct.
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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.
Using the factor theorem, it is found that the roots are:
d. 6 with multiplicity 2
f. -2 with multiplicity 2
The Factor Theorem states that if a polynomial has roots [tex]x_1, x_2, ..., x_n[/tex], each with multiplicity [tex]n_1, n_2, ..., n_n[/tex], the polynomial is written as:
[tex]p(x) = (x - x_1)^{n_1}(x - x_2)^{n_2}...(x - x_n)^{n_n}[/tex]
In this problem, the polynomial is:
[tex]f(x) = (x - 6)^2(x + 2)^2[/tex]
Thus:
Options d and f, thus:
A similar problem is given at https://brainly.com/question/24380382
