Question Find the true statements about the polynomial $f(x) = x^{2} - 12x + 36$ . Options: A) Two distinct linear factors, B) Roots at $x = \pm 6$ , C) At least one complex root, D) Satisfies Fundamental Theorem of Algebra.

Studdy Solution

STEP 1

Assumptions
1. The function given is $f(x) = x^2 - 12x + 36$ .
2. We need to determine the truth of statements A, B, C, and D regarding the function $f(x)$ .

STEP 2

To determine if statement A is true, we need to factor the polynomial if possible. We look for two numbers that multiply to the constant term (36) and add up to the coefficient of the linear term (-12).

STEP 3

We find that the numbers -6 and -6 multiply to 36 and add up to -12.

STEP 4

Factor the polynomial using these numbers.
$f(x) = (x - 6)(x - 6)$

STEP 5

Since the factors are the same, we can write the polynomial as the square of a binomial.
$f(x) = (x - 6)^2$

STEP 6

Determine the truth of statement A: The polynomial which defines $f(x)$ has two distinct linear factors.
Since the factors are not distinct (they are the same factor repeated), statement A is false.

STEP 7

To determine if statement B is true, we need to find the roots of the function $f(x)$ . The roots are the values of $x$ for which $f(x) = 0$ .

STEP 8

Set the factored form of $f(x)$ equal to zero and solve for $x$ .
$(x - 6)^2 = 0$

STEP 9

Take the square root of both sides.
$x - 6 = 0$

STEP 10

Solve for $x$ .
$x = 6$

STEP 11

Determine the truth of statement B: The function $f(x)$ has roots at $x = \pm 6$ .
Since the only root is $x = 6$ , and there is no root at $x = -6$ , statement B is false.

STEP 12

To determine if statement C is true, we need to consider the nature of the roots of the polynomial. A quadratic polynomial with real coefficients will have complex roots if the discriminant is negative.

STEP 13

Calculate the discriminant of the quadratic polynomial $f(x) = x^2 - 12x + 36$ .
$\text{Discriminant} = b^2 - 4ac$

STEP 14

Plug in the values of $a = 1$ , $b = -12$ , and $c = 36$ into the discriminant formula.
$\text{Discriminant} = (-12)^2 - 4(1)(36)$

STEP 15

Calculate the discriminant.
$\text{Discriminant} = 144 - 144 = 0$

STEP 16

Determine the truth of statement C: The function $f(x)$ has at least one complex root.
Since the discriminant is zero, the roots are real and not complex. Therefore, statement C is false.

STEP 17

To determine if statement D is true, we need to understand the Fundamental Theorem of Algebra, which states that every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

STEP 18

Since $f(x)$ is a degree 2 polynomial, it must have 2 roots, which can be real or complex.

STEP 19

Determine the truth of statement D: The polynomial which defines $f(x)$ satisfies the Fundamental Theorem of Algebra.
Since $f(x)$ has exactly 2 roots (both are $x = 6$ ), it satisfies the Fundamental Theorem of Algebra. Therefore, statement D is true.
The true statements about $f(x) = x^2 - 12x + 36$ are:
D The polynomial which defines $f(x)$ satisfies the Fundamental Theorem of Algebra.

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Question Find the true statements about the polynomial f(x)=x2−12x+36f(x) = x^{2} - 12x + 36f(x)=x2−12x+36. Options: A) Two distinct linear factors, B) Roots at x=±6x = \pm 6x=±6, C) At least one complex root, D) Satisfies Fundamental Theorem of Algebra.

Studdy Solution

STEP 1

Assumptions1. The function given is f(x)=x2−12x+36f(x) = x^2 - 12x + 36f(x)=x2−12x+36.2. We need to determine the truth of statements A, B, C, and D regarding the function f(x)f(x)f(x).

STEP 2

To determine if statement A is true, we need to factor the polynomial if possible. We look for two numbers that multiply to the constant term (36) and add up to the coefficient of the linear term (-12).

STEP 3

We find that the numbers -6 and -6 multiply to 36 and add up to -12.

STEP 4

Factor the polynomial using these numbers.f(x)=(x−6)(x−6)f(x) = (x - 6)(x - 6)f(x)=(x−6)(x−6)

STEP 5

Since the factors are the same, we can write the polynomial as the square of a binomial.f(x)=(x−6)2f(x) = (x - 6)^2f(x)=(x−6)2

STEP 6

Determine the truth of statement A: The polynomial which defines f(x)f(x)f(x) has two distinct linear factors.Since the factors are not distinct (they are the same factor repeated), statement A is false.

STEP 7

To determine if statement B is true, we need to find the roots of the function f(x)f(x)f(x). The roots are the values of xxx for which f(x)=0f(x) = 0f(x)=0.

STEP 8

Set the factored form of f(x)f(x)f(x) equal to zero and solve for xxx.(x−6)2=0(x - 6)^2 = 0(x−6)2=0

STEP 9

Take the square root of both sides.x−6=0x - 6 = 0x−6=0

STEP 10

Solve for xxx.x=6x = 6x=6

STEP 11

Determine the truth of statement B: The function f(x)f(x)f(x) has roots at x=±6x = \pm 6x=±6.Since the only root is x=6x = 6x=6, and there is no root at x=−6x = -6x=−6, statement B is false.

STEP 12

To determine if statement C is true, we need to consider the nature of the roots of the polynomial. A quadratic polynomial with real coefficients will have complex roots if the discriminant is negative.

STEP 13

Calculate the discriminant of the quadratic polynomial f(x)=x2−12x+36f(x) = x^2 - 12x + 36f(x)=x2−12x+36.Discriminant=b2−4ac\text{Discriminant} = b^2 - 4acDiscriminant=b2−4ac

STEP 14

Plug in the values of a=1a = 1a=1, b=−12b = -12b=−12, and c=36c = 36c=36 into the discriminant formula.Discriminant=(−12)2−4(1)(36)\text{Discriminant} = (-12)^2 - 4(1)(36)Discriminant=(−12)2−4(1)(36)

STEP 15

Calculate the discriminant.Discriminant=144−144=0\text{Discriminant} = 144 - 144 = 0Discriminant=144−144=0

STEP 16

Determine the truth of statement C: The function f(x)f(x)f(x) has at least one complex root.Since the discriminant is zero, the roots are real and not complex. Therefore, statement C is false.

STEP 17

To determine if statement D is true, we need to understand the Fundamental Theorem of Algebra, which states that every non-zero, single-variable, degree nnn polynomial with complex coefficients has, counted with multiplicity, exactly nnn complex roots.

STEP 18

Since f(x)f(x)f(x) is a degree 2 polynomial, it must have 2 roots, which can be real or complex.

STEP 19

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Question Find the true statements about the polynomial $f(x) = x^{2} - 12x + 36$ . Options: A) Two distinct linear factors, B) Roots at $x = \pm 6$ , C) At least one complex root, D) Satisfies Fundamental Theorem of Algebra.

Assumptions
1. The function given is $f(x) = x^2 - 12x + 36$ .
2. We need to determine the truth of statements A, B, C, and D regarding the function $f(x)$ .

Factor the polynomial using these numbers.
$f(x) = (x - 6)(x - 6)$

Since the factors are the same, we can write the polynomial as the square of a binomial.
$f(x) = (x - 6)^2$

Determine the truth of statement A: The polynomial which defines $f(x)$ has two distinct linear factors.
Since the factors are not distinct (they are the same factor repeated), statement A is false.

To determine if statement B is true, we need to find the roots of the function $f(x)$ . The roots are the values of $x$ for which $f(x) = 0$ .

Set the factored form of $f(x)$ equal to zero and solve for $x$ .
$(x - 6)^2 = 0$

Take the square root of both sides.
$x - 6 = 0$

Solve for $x$ .
$x = 6$

Determine the truth of statement B: The function $f(x)$ has roots at $x = \pm 6$ .
Since the only root is $x = 6$ , and there is no root at $x = -6$ , statement B is false.

Calculate the discriminant of the quadratic polynomial $f(x) = x^2 - 12x + 36$ .
$\text{Discriminant} = b^2 - 4ac$

Plug in the values of $a = 1$ , $b = -12$ , and $c = 36$ into the discriminant formula.
$\text{Discriminant} = (-12)^2 - 4(1)(36)$

Calculate the discriminant.
$\text{Discriminant} = 144 - 144 = 0$

Determine the truth of statement C: The function $f(x)$ has at least one complex root.
Since the discriminant is zero, the roots are real and not complex. Therefore, statement C is false.

To determine if statement D is true, we need to understand the Fundamental Theorem of Algebra, which states that every non-zero, single-variable, degree $n$ polynomial with complex coefficients has, counted with multiplicity, exactly $n$ complex roots.

Since $f(x)$ is a degree 2 polynomial, it must have 2 roots, which can be real or complex.