Find solutions to your problems with the help of IDNLearn.com's expert community. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.
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 presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.