Question

the function f(x) = 12x³ - 20x² + x + 3 has at least one rational root. use the rational root theorem to find that root, then proceed to find all complex roots. (note: roots may be integer, rational, irrational, and/or complex.) answer attempt 1 out of 2 there is one root :

Explanation:

Step1: Apply Rational Root Theorem

The Rational Root Theorem states that any possible rational root, \( \frac{p}{q} \), of a polynomial \( a_nx^n + a_{n - 1}x^{n - 1}+\dots+a_1x + a_0 \) is a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \). For \( f(x)=12x^3 - 20x^2 + x + 3 \), the constant term \( a_0 = 3 \) and the leading coefficient \( a_3=12 \). The factors of \( 3 \) are \( \pm1,\pm3 \) and the factors of \( 12 \) are \( \pm1,\pm2,\pm3,\pm4,\pm6,\pm12 \). So the possible rational roots are \( \pm1,\pm3,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{1}{4},\pm\frac{3}{4},\pm\frac{1}{6},\pm\frac{1}{12},\pm\frac{3}{6}=\pm\frac{1}{2},\pm\frac{3}{12}=\pm\frac{1}{4} \) (removing duplicates).

Step2: Test Possible Rational Roots

Test \( x = 1 \): \( f(1)=12 - 20 + 1+ 3=-4
eq0 \)
Test \( x=-1 \): \( f(-1)=-12 - 20 - 1 + 3=-30
eq0 \)
Test \( x = 3 \): \( f(3)=12\times27-20\times9 + 3+ 3=324 - 180+6 = 150
eq0 \)
Test \( x=\frac{1}{2} \): \( f(\frac{1}{2})=12\times\frac{1}{8}-20\times\frac{1}{4}+\frac{1}{2}+3=\frac{3}{2}-5+\frac{1}{2}+3=-1
eq0 \)
Test \( x=\frac{3}{2} \): \( f(\frac{3}{2})=12\times\frac{27}{8}-20\times\frac{9}{4}+\frac{3}{2}+3=\frac{81}{2}-45+\frac{3}{2}+3=\frac{81 + 3}{2}-42=42 - 42=0 \)
So \( x=\frac{3}{2} \) is a rational root.

Step3: Factor the Polynomial

Since \( x = \frac{3}{2} \) is a root, \( (2x - 3) \) is a factor. We perform polynomial division or use synthetic division. Using synthetic division with root \( \frac{3}{2} \):
The coefficients are \( 12,-20,1,3 \)
Bring down \( 12 \). Multiply \( 12\times\frac{3}{2}=18 \). Add to \( -20 \): \( -20 + 18=-2 \)
Multiply \( -2\times\frac{3}{2}=-3 \). Add to \( 1 \): \( 1-3=-2 \)
Multiply \( -2\times\frac{3}{2}=-3 \). Add to \( 3 \): \( 3 - 3=0 \)
So the quotient polynomial is \( 12x^2-2x - 2 \). We can factor out a \( 2 \): \( 2(6x^2 - x - 1) \). Then factor \( 6x^2 - x - 1=(3x + 1)(2x - 1) \). So \( f(x)=(2x - 3)\times2\times(3x + 1)(2x - 1)=2(2x - 3)(3x + 1)(2x - 1) \)

Step4: Find All Roots

Set each factor equal to zero:

• \( 2x - 3 = 0\Rightarrow x=\frac{3}{2} \)
• \( 3x + 1 = 0\Rightarrow x=-\frac{1}{3} \)
• \( 2x - 1 = 0\Rightarrow x=\frac{1}{2} \)

Answer:

The rational root we found first is \( \frac{3}{2} \), and all complex roots (which are real here) are \( \frac{3}{2},-\frac{1}{3},\frac{1}{2} \). But for the first root (the one we found using the Rational Root Theorem), the answer is \( \frac{3}{2} \)