At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Our Q&A platform offers a seamless experience for finding reliable answers from experts in various disciplines. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
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.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.