Question

use synthetic division to find the real zeroes of $f(x)=2x^3 + 11x^2 + 12x - 9$.(1 point)\\(\bigcirc -1,\frac{3}{2},9\\)\\(\bigcirc -3,\frac{1}{2}\\)\\(\bigcirc -\frac{1}{2},1,3\\)\\(\bigcirc -\frac{9}{2},3\\)

Explanation:

Step 1: Find possible rational roots

Using the Rational Root Theorem, the possible rational roots are factors of the constant term over factors of the leading coefficient. For \( f(x) = 2x^3 + 11x^2 + 12x - 9 \), the constant term is \(-9\) and the leading coefficient is \(2\). So possible roots are \(\pm1, \pm3, \pm9, \pm\frac{1}{2}, \pm\frac{3}{2}, \pm\frac{9}{2}\).

Step 2: Test possible roots with synthetic division

Let's test \( x = \frac{1}{2} \):
The coefficients are \(2, 11, 12, -9\).
Bring down the \(2\). Multiply \(2\times\frac{1}{2}=1\), add to \(11\): \(11 + 1 = 12\). Multiply \(12\times\frac{1}{2}=6\), add to \(12\): \(12 + 6 = 18\). Multiply \(18\times\frac{1}{2}=9\), add to \(-9\): \(-9 + 9 = 0\). So \( x = \frac{1}{2} \) is a root? Wait, no, wait, maybe I made a mistake. Wait, let's test \( x = -3 \):
Coefficients: \(2, 11, 12, -9\)
Bring down \(2\). Multiply \(2\times(-3)= -6\), add to \(11\): \(11 - 6 = 5\). Multiply \(5\times(-3)= -15\), add to \(12\): \(12 - 15 = -3\). Multiply \(-3\times(-3)= 9\), add to \(-9\): \(-9 + 9 = 0\). So \( x = -3 \) is a root. Now, the quotient polynomial is \(2x^2 + 5x - 3\).

Step 3: Factor the quadratic

Factor \(2x^2 + 5x - 3\). We need two numbers that multiply to \(2\times(-3)= -6\) and add to \(5\). Those numbers are \(6\) and \(-1\). So rewrite as \(2x^2 + 6x - x - 3 = 2x(x + 3) - 1(x + 3) = (2x - 1)(x + 3)\). Wait, but we already have \(x = -3\) as a root. Wait, no, the quadratic is \(2x^2 + 5x - 3\), setting it to zero: \(2x^2 + 5x - 3 = 0\). Using quadratic formula \(x = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm 7}{4}\). So \(x = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}\) or \(x = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\). Wait, but that would mean the roots are \(-3, \frac{1}{2}\) (and \(-3\) is a repeated root? No, wait, the original polynomial is cubic, so three roots. Wait, maybe I made a mistake in synthetic division. Wait, let's try \(x = \frac{1}{2}\) again. Wait, no, when we tested \(x = -3\), we got a zero remainder, so the polynomial factors as \((x + 3)(2x^2 + 5x - 3)\). Then factoring \(2x^2 + 5x - 3\) gives \((2x - 1)(x + 3)\)? Wait, no, \((2x - 1)(x + 3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3\), correct. So the roots are \(x = -3\) (from \(x + 3\)) and \(x = \frac{1}{2}\) (from \(2x - 1\)), but wait, that's a quadratic, so the cubic has a root at \(x = -3\) (with multiplicity 2?) No, wait, the original polynomial is \(2x^3 + 11x^2 + 12x - 9\). Let's check \(x = -3\): \(2(-27) + 11(9) + 12(-3) - 9 = -54 + 99 - 36 - 9 = 0\). \(x = \frac{1}{2}\): \(2(\frac{1}{8}) + 11(\frac{1}{4}) + 12(\frac{1}{2}) - 9 = \frac{1}{4} + \frac{11}{4} + 6 - 9 = \frac{12}{4} - 3 = 3 - 3 = 0\). Wait, but then where is the third root? Wait, maybe I made a mistake in synthetic division. Wait, let's try \(x = \frac{1}{2}\) again. Wait, no, the cubic has three roots. Wait, maybe my initial synthetic division was wrong. Wait, let's try \(x = -3\) again. Coefficients: 2, 11, 12, -9. Bring down 2. Multiply by -3: -6. Add to 11: 5. Multiply by -3: -15. Add to 12: -3. Multiply by -3: 9. Add to -9: 0. Correct. So the quotient is \(2x^2 + 5x - 3\). Now, let's factor \(2x^2 + 5x - 3\) again. Wait, maybe I should use synthetic division on the quadratic? No, it's a quadratic. Wait, but the options include \(-3, \frac{1}{2}\) as an option (the second option: \(-3, \frac{1}{2}\)). Let's check the options. The options are:

1. \(-1, \frac{3}{2}, 9\)
2. \(-3, \frac{1}{2}\)
3. \(-\frac{1}{2}, 1, 3\)
4. \(-\frac{9}{2}, 3\)

Wait, but a cubic should have three real roots (since it's a cubic with real coefficients). Wait, maybe I made a mistake in the synthetic division. Wait, let's try \(x = \frac{1}{2}\) again. Wait, no, when we plug \(x = \frac{1}{2}\) into \(f(x)\): \(2(\frac{1}{2})^3 + 11(\frac{1}{2})^2 + 12(\frac{1}{2}) - 9 = 2(\frac{1}{8}) + 11(\frac{1}{4}) + 6 - 9 = \frac{1}{4} + \frac{11}{4} - 3 = \frac{12}{4} - 3 = 3 - 3 = 0\). So \(x = \frac{1}{2}\) is a root. \(x = -3\): \(2(-27) + 11(9) + 12(-3) - 9 = -54 + 99 - 36 - 9 = 0\). Wait, but then we have two roots, but it's a cubic. Wait, maybe there's a repeated root? Wait, the quadratic \(2x^2 + 5x - 3\) has roots \(-3\) and \(\frac{1}{2}\), so the cubic is \((x + 3)(2x^2 + 5x - 3) = (x + 3)(2x - 1)(x + 3) = (x + 3)^2(2x - 1)\). So the roots are \(-3\) (multiplicity 2) and \(\frac{1}{2}\). But the options have \(-3, \frac{1}{2}\) as the second option. Let's check the options again. The second option is \(-3, \frac{1}{2}\), which are two roots, but maybe the question considers the distinct real roots? Wait, no, a cubic must have three real roots (counting multiplicity). But maybe the options have a typo, or I made a mistake. Wait, let's check the first option: \(-1, \frac{3}{2}, 9\). Plug \(x = -1\): \(2(-1)^3 + 11(-1)^2 + 12(-1) - 9 = -2 + 11 - 12 - 9 = -12 eq 0\). \(x = \frac{3}{2}\): \(2(\frac{27}{8}) + 11(\frac{9}{4}) + 12(\frac{3}{2}) - 9 = \frac{27}{4} + \frac{99}{4} + 18 - 9 = \frac{126}{4} + 9 = 31.5 + 9 = 40.5 eq 0\). \(x = 9\): way too big, not zero. Third option: \(-\frac{1}{2}\): \(2(-\frac{1}{2})^3 + 11(-\frac{1}{2})^2 + 12(-\frac{1}{2}) - 9 = 2(-\frac{1}{8}) + 11(\frac{1}{4}) - 6 - 9 = -\frac{1}{4} + \frac{11}{4} - 15 = \frac{10}{4} - 15 = 2.5 - 15 = -12.5 eq 0\). \(x = 1\): \(2 + 11 + 12 - 9 = 16 eq 0\). \(x = 3\): \(2(27) + 11(9) + 12(3) - 9 = 54 + 99 + 36 - 9 = 180 eq 0\). Fourth option: \(-\frac{9}{2}\): \(2(-\frac{729}{8}) + 11(\frac{81}{4}) + 12(-\frac{9}{2}) - 9 = -\frac{729}{4} + \frac{891}{4} - 54 - 9 = \frac{162}{4} - 63 = 40.5 - 63 = -22.5 eq 0\). \(x = 3\): as above, not zero. So the only option with roots that work is the second option: \(-3, \frac{1}{2}\). Wait, but why is it a cubic? Maybe there's a repeated root at \(-3\), so the real zeros are \(-3\) (with multiplicity 2) and \(\frac{1}{2}\), but the option lists \(-3, \frac{1}{2}\) as the real zeros (ignoring multiplicity). So that's the correct option.

Answer:

B. \(-3, \frac{1}{2}\)