Question

which of the following is a possible root of \\( f(x) = 2x^2 - 5x + 12 \\)?\
\\( \frac{1}{6} \\)\
\\( \frac{3}{2} \\)\
\\( \frac{1}{3} \\)\
24

Explanation:

Step1: Recall Rational Root Theorem

The Rational Root Theorem states that any possible rational root of a polynomial \( f(x) = a_nx^n + a_{n - 1}x^{n - 1}+\cdots+a_1x + a_0\) (where \(a_n
eq0\)) is of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\).

For the polynomial \(f(x)=2x^2 - 5x + 12\), the leading coefficient \(a_n = 2\) (factors of \(2\) are \(\pm1,\pm2\)) and the constant term \(a_0=12\) (factors of \(12\) are \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12\)). So the possible rational roots are \(\pm\frac{1}{1},\pm\frac{2}{1},\pm\frac{3}{1},\pm\frac{4}{1},\pm\frac{6}{1},\pm\frac{12}{1},\pm\frac{1}{2},\pm\frac{2}{2},\pm\frac{3}{2},\pm\frac{4}{2},\pm\frac{6}{2},\pm\frac{12}{2}\). Simplifying, the possible rational roots are \(\pm1,\pm2,\pm3,\pm4,\pm6,\pm12,\pm\frac{1}{2},\pm\frac{3}{2}\).

Step2: Check each option

• Option \(\frac{1}{6}\): \(6\) is not a factor of the leading coefficient \(2\), so \(\frac{1}{6}\) is not a possible rational root.
• Option \(\frac{3}{2}\): \(3\) is a factor of \(12\) (constant term) and \(2\) is a factor of \(2\) (leading coefficient), so \(\frac{3}{2}\) is a possible rational root (we can verify by plugging into the polynomial: \(f(\frac{3}{2})=2\times(\frac{3}{2})^2-5\times\frac{3}{2}+12=2\times\frac{9}{4}-\frac{15}{2}+12=\frac{9}{2}-\frac{15}{2}+12=\frac{9 - 15}{2}+12=\frac{- 6}{2}+12=- 3 + 12 = 9

eq0\), but it is still a possible rational root as per the Rational Root Theorem (the theorem gives possible roots, not necessarily actual roots)).

• Option \(\frac{1}{3}\): \(3\) is not a factor of the leading coefficient \(2\), so \(\frac{1}{3}\) is not a possible rational root.
• Option \(24\): \(24\) is not of the form \(\frac{p}{q}\) where \(p\) is a factor of \(12\) and \(q\) is a factor of \(2\) (since \(24=\frac{24}{1}\), but \(24\) is not a factor of \(12\)), so \(24\) is not a possible rational root.

Answer:

\(\frac{3}{2}\) (i.e., the option \(\boldsymbol{\frac{3}{2}}\))