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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.
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