Question

the function f(x) = 2x³ + 3x² - 11x + 6 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 possible rational roots of a polynomial \(a_nx^n + \dots + a_1x + a_0\) are of the form \(\frac{p}{q}\), where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. For \(f(x)=2x^3 + 3x^2 - 11x + 6\), the constant term is \(6\) (factors: \(\pm1, \pm2, \pm3, \pm6\)) and the leading coefficient is \(2\) (factors: \(\pm1, \pm2\)). So possible rational roots are \(\pm1, \pm2, \pm3, \pm6, \pm\frac{1}{2}, \pm\frac{3}{2}\).

Test \(x = 1\): \(f(1)=2 + 3 - 11 + 6 = 0\). So \(x = 1\) is a root.

Step2: Factor the Polynomial

Since \(x = 1\) is a root, \((x - 1)\) is a factor. Use polynomial division or synthetic division to divide \(f(x)\) by \((x - 1)\). Using synthetic division:

\[ \begin{array}{r|rrrr} 1 & 2 & 3 & -11 & 6 \\ & & 2 & 5 & -6 \\ \hline & 2 & 5 & -6 & 0 \\ \end{array} \]

So \(f(x)=(x - 1)(2x^2 + 5x - 6)\)? Wait, no, wait, the quotient is \(2x^2 + 5x - 6\)? Wait, no, wait, let's recalculate. Wait, \(2x^3 + 3x^2 - 11x + 6\) divided by \(x - 1\):

\(2x^3\div x = 2x^2\), multiply \((x - 1)\) by \(2x^2\) to get \(2x^3 - 2x^2\). Subtract from \(2x^3 + 3x^2\) to get \(5x^2\). Bring down \(-11x\): \(5x^2 - 11x\). Divide \(5x^2\) by \(x\) to get \(5x\), multiply \((x - 1)\) by \(5x\) to get \(5x^2 - 5x\). Subtract to get \(-6x\). Bring down \(6\): \(-6x + 6\). Divide \(-6x\) by \(x\) to get \(-6\), multiply \((x - 1)\) by \(-6\) to get \(-6x + 6\). Subtract to get \(0\). So the quotient is \(2x^2 + 5x - 6\)? Wait, no, wait, that can't be. Wait, maybe I made a mistake. Wait, let's try \(x = \frac{1}{2}\): \(f(\frac{1}{2})=2(\frac{1}{8}) + 3(\frac{1}{4}) - 11(\frac{1}{2}) + 6=\frac{1}{4}+\frac{3}{4}-\frac{11}{2}+6 = 1 - \frac{11}{2}+6=\frac{2 - 11 + 12}{2}=\frac{3}{2} eq0\). Wait, \(x = 2\): \(f(2)=16 + 12 - 22 + 6 = 12 eq0\). \(x = -1\): \(f(-1)=-2 + 3 + 11 + 6 = 18 eq0\). \(x = 3\): \(f(3)=54 + 27 - 33 + 6 = 54 eq0\). \(x = -2\): \(f(-2)=-16 + 12 + 22 + 6 = 24 eq0\). \(x = \frac{3}{2}\): \(f(\frac{3}{2})=2(\frac{27}{8}) + 3(\frac{9}{4}) - 11(\frac{3}{2}) + 6=\frac{27}{4}+\frac{27}{4}-\frac{33}{2}+6=\frac{54}{4}-\frac{66}{4}+\frac{24}{4}=\frac{54 - 66 + 24}{4}=\frac{12}{4}=3 eq0\). Wait, earlier when I tested \(x = 1\), I think I miscalculated. Wait, \(f(1)=2(1)^3 + 3(1)^2 - 11(1) + 6 = 2 + 3 - 11 + 6 = 0\). Oh, right, that's correct. So then the quotient is \(2x^2 + 5x - 6\)? Wait, no, wait, \(2x^3 + 3x^2 - 11x + 6=(x - 1)(2x^2 + 5x - 6)\)? Wait, no, let's multiply \((x - 1)(2x^2 + 5x - 6)=2x^3 + 5x^2 - 6x - 2x^2 - 5x + 6=2x^3 + 3x^2 - 11x + 6\). Yes, that's correct. Now, let's factor \(2x^2 + 5x - 6\) or use quadratic formula. Wait, maybe another root. Wait, maybe I made a mistake in possible roots. Wait, let's try \(x = \frac{3}{2}\) again. Wait, \(f(\frac{3}{2})=2(\frac{27}{8}) + 3(\frac{9}{4}) - 11(\frac{3}{2}) + 6=\frac{27}{4}+\frac{27}{4}-\frac{33}{2}+6=\frac{54}{4}-\frac{66}{4}+\frac{24}{4}=\frac{12}{4}=3\). No. Wait, \(x = -3\): \(f(-3)=2(-27) + 3(9) - 11(-3) + 6=-54 + 27 + 33 + 6=12 eq0\). Wait, maybe \(x = \frac{1}{2}\) was miscalculated. \(f(\frac{1}{2})=2(\frac{1}{8}) + 3(\frac{1}{4}) - 11(\frac{1}{2}) + 6=\frac{1}{4}+\frac{3}{4}-\frac{11}{2}+6=1 - \frac{11}{2}+6=\frac{2 - 11 + 12}{2}=\frac{3}{2}\). No. Wait, maybe the first root is \(x = 1\), then we can also try \(x = \frac{3}{2}\) again? Wait, no, let's use the quadratic formula on \(2x^2 + 5x - 6\). The quadratic formula is \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 5\), \(c = -6\). So \(x=\frac{-5\pm\sqrt{25 + 48}}{4}=\frac{-5\pm\sqrt{73}}{4}\). But that's irrational. Wait, but maybe I made a mistake in the synthetic division. Wait, let's do synthetic division again with root \(x = 1\):

Coefficients: \(2\) (x³), \(3\) (x²), \(-11\) (x), \(6\) (constant)

Bring down \(2\). Multiply by \(1\): \(2\). Add to next coefficient: \(3 + 2 = 5\). Multiply by \(1\): \(5\). Add to next coefficient: \(-11 + 5 = -6\). Multiply by \(1\): \(-6\). Add to last coefficient: \(6 + (-6) = 0\). So the quotient polynomial is \(2x^2 + 5x - 6\). Wait, but maybe there's another rational root. Let's try \(x = -3\): \(f(-3)=2(-27) + 3(9) - 11(-3) + 6=-54 + 27 + 33 + 6=12 eq0\). \(x = 6\): \(f(6)=2(216) + 3(36) - 11(6) + 6=432 + 108 - 66 + 6=480 eq0\). \(x = -6\): \(f(-6)=2(-216) + 3(36) - 11(-6) + 6=-432 + 108 + 66 + 6=-252 eq0\). Wait, maybe I made a mistake in the polynomial. Wait, the polynomial is \(2x^3 + 3x^2 - 11x + 6\). Let's try \(x = \frac{1}{2}\) again: \(f(\frac{1}{2})=2(\frac{1}{8}) + 3(\frac{1}{4}) - 11(\frac{1}{2}) + 6=\frac{1}{4}+\frac{3}{4}-\frac{11}{2}+6=1 - \frac{11}{2}+6=\frac{2 - 11 + 12}{2}=\frac{3}{2}\). No. Wait, \(x = 2\): \(f(2)=16 + 12 - 22 + 6=12\). \(x = -1\): \(f(-1)=-2 + 3 + 11 + 6=18\). Wait, maybe the first root is \(x = 1\), and then we can factor the quadratic \(2x^2 + 5x - 6\) or use quadratic formula. But the problem says "at least one rational root", so \(x = 1\) is a rational root. Wait, but maybe I missed another rational root. Let's try \(x = \frac{3}{2}\) again. Wait, \(f(\frac{3}{2})=2(\frac{27}{8}) + 3(\frac{9}{4}) - 11(\frac{3}{2}) + 6=\frac{27}{4}+\frac{27}{4}-\frac{33}{2}+6=\frac{54}{4}-\frac{66}{4}+\frac{24}{4}=\frac{12}{4}=3\). No. Wait, \(x = -2\): \(f(-2)=2(-8) + 3(4) - 11(-2) + 6=-16 + 12 + 22 + 6=24\). Wait, maybe the polynomial can be factored as \((x - 1)(2x - 1)(x + 3)\)? Let's check: \((x - 1)(2x - 1)(x + 3)=(x - 1)(2x^2 + 6x - x - 3)=(x - 1)(2x^2 + 5x - 3)=2x^3 + 5x^2 - 3x - 2x^2 - 5x + 3=2x^3 + 3x^2 - 8x + 3\). No, that's not the polynomial. Wait, maybe I made a mistake in the synthetic division. Wait, the coefficients after division are \(2\), \(5\), \(-6\), so the quadratic is \(2x^2 + 5x - 6\). The discriminant of this quadratic is \(b^2 - 4ac = 25 - 4(2)(-6)=25 + 48=73\), which is positive, so the roots are \(\frac{-5\pm\sqrt{73}}{4}\), which are irrational. But the problem says "at least one rational root", so \(x = 1\) is the rational root. Wait, but maybe I made a mistake in testing \(x = 1\). Wait, \(f(1)=2 + 3 - 11 + 6=0\), that's correct. So the rational root is \(x = 1\), and the other roots are \(\frac{-5 + \sqrt{73}}{4}\) and \(\frac{-5 - \sqrt{73}}{4}\). But the problem says "There is one root ~ :", maybe it's asking for the rational root first. So the rational root is \(1\).

Answer:

\(1\)