Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To determine which statement is true about the polynomial [tex]\( f(x) = 12x^3 - 5x^2 + 6x + 9 \)[/tex], we will use the Rational Root Theorem.
The Rational Root Theorem states that any rational root of a polynomial [tex]\( f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)[/tex] is of the form [tex]\(\frac{p}{q}\)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term [tex]\( a_0 \)[/tex] and [tex]\( q \)[/tex] is a factor of the leading coefficient [tex]\( a_n \)[/tex].
For the given polynomial [tex]\( f(x) = 12x^3 - 5x^2 + 6x + 9 \)[/tex]:
- The constant term [tex]\( a_0 \)[/tex] is 9.
- The leading coefficient [tex]\( a_n \)[/tex] is 12.
First, we identify the factors of the constant term (9) and the leading coefficient (12).
Factors of 9:
[tex]\[ 1, 3, 9, -1, -3, -9 \][/tex]
Factors of 12:
[tex]\[ 1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12 \][/tex]
According to the Rational Root Theorem, any rational root [tex]\( \frac{p}{q} \)[/tex] of [tex]\( f(x) \)[/tex] must be a factor of 9 divided by a factor of 12. Let's list all possible combinations of [tex]\( \frac{p}{q} \)[/tex] that meet these criteria.
Possible rational roots:
[tex]\[ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}, \frac{3}{1}, \frac{3}{2}, \frac{3}{3}, \frac{3}{4}, \frac{3}{6}, \frac{3}{12}, \frac{9}{1}, \frac{9}{2}, \frac{9}{3}, \frac{9}{4}, \frac{9}{6}, \frac{9}{12}, \frac{-1}{1}, \frac{-1}{2}, \frac{-1}{3}, \frac{-1}{4}, \frac{-1}{6}, \frac{-1}{12}, \frac{-3}{1}, \frac{-3}{2}, \frac{-3}{3}, \frac{-3}{4}, \frac{-3}{6}, \frac{-3}{12}, \frac{-9}{1}, \frac{-9}{2}, \frac{-9}{3}, \frac{-9}{4}, \frac{-9}{6}, \frac{-9}{12} \][/tex]
After removing the duplicates, the unique possible rational roots are as follows:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{6}, \pm \frac{3}{12}, \pm 9, \pm \frac{9}{2}, \pm \frac{9}{3}, \pm \frac{9}{4}, \pm \frac{9}{6}, \pm \frac{9}{12} \][/tex]
Therefore, the correct statement is:
Any rational root of [tex]\( f(x) \)[/tex] is a factor of 9 divided by a factor of 12.
The Rational Root Theorem states that any rational root of a polynomial [tex]\( f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)[/tex] is of the form [tex]\(\frac{p}{q}\)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term [tex]\( a_0 \)[/tex] and [tex]\( q \)[/tex] is a factor of the leading coefficient [tex]\( a_n \)[/tex].
For the given polynomial [tex]\( f(x) = 12x^3 - 5x^2 + 6x + 9 \)[/tex]:
- The constant term [tex]\( a_0 \)[/tex] is 9.
- The leading coefficient [tex]\( a_n \)[/tex] is 12.
First, we identify the factors of the constant term (9) and the leading coefficient (12).
Factors of 9:
[tex]\[ 1, 3, 9, -1, -3, -9 \][/tex]
Factors of 12:
[tex]\[ 1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12 \][/tex]
According to the Rational Root Theorem, any rational root [tex]\( \frac{p}{q} \)[/tex] of [tex]\( f(x) \)[/tex] must be a factor of 9 divided by a factor of 12. Let's list all possible combinations of [tex]\( \frac{p}{q} \)[/tex] that meet these criteria.
Possible rational roots:
[tex]\[ \frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}, \frac{3}{1}, \frac{3}{2}, \frac{3}{3}, \frac{3}{4}, \frac{3}{6}, \frac{3}{12}, \frac{9}{1}, \frac{9}{2}, \frac{9}{3}, \frac{9}{4}, \frac{9}{6}, \frac{9}{12}, \frac{-1}{1}, \frac{-1}{2}, \frac{-1}{3}, \frac{-1}{4}, \frac{-1}{6}, \frac{-1}{12}, \frac{-3}{1}, \frac{-3}{2}, \frac{-3}{3}, \frac{-3}{4}, \frac{-3}{6}, \frac{-3}{12}, \frac{-9}{1}, \frac{-9}{2}, \frac{-9}{3}, \frac{-9}{4}, \frac{-9}{6}, \frac{-9}{12} \][/tex]
After removing the duplicates, the unique possible rational roots are as follows:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 3, \pm \frac{3}{2}, \pm \frac{3}{4}, \pm \frac{3}{6}, \pm \frac{3}{12}, \pm 9, \pm \frac{9}{2}, \pm \frac{9}{3}, \pm \frac{9}{4}, \pm \frac{9}{6}, \pm \frac{9}{12} \][/tex]
Therefore, the correct statement is:
Any rational root of [tex]\( f(x) \)[/tex] is a factor of 9 divided by a factor of 12.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.