Question 3 (2 points) Factor completely 12x^3-40x^2-27x+90 A (4x^2-9)(3x+10) (4x-3)(3x-10)(x+3) (2x-3)(6x^2-11x-30) (2x+3)(2x-3)(3x-10)

To factor the polynomial \(12x^3 - 40x^2 - 27x + 90\) completely, we can use the Rational Root Theorem to find possible rational roots and then perform polynomial division or synthetic division.<br /><br />First, let's list the possible rational roots. The Rational Root Theorem states that any rational root of the polynomial equation \(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0\) is a fraction \(\frac{p}{q}\), where \(p\) is a factor of the constant term \(a_0\) and \(q\) is a factor of the leading coefficient \(a_n\).<br /><br />For the polynomial \(12x^3 - 40x^2 - 27x + 90\):<br />- The constant term \(a_0 = 90\)<br />- The leading coefficient \(a_n = 12\)<br /><br />Factors of 90: \(\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 9, \pm 10, \pm 15, \pm 18, \pm 30, \pm 45, \pm 90\)<br />Factors of 12: \(\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12\)<br /><br />Possible rational roots are:<br />\[<br />\pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{4}, \pm \frac{1}{6}, \pm \frac{1}{12}, \pm 2, \pm 3, \pm 5pm 6, \pm 9, \pm 10, \pm 15, \pm 18, \pm 30, \pm 45, \pm 90<br />\]<br /><br />We will test these possible roots by substituting them into the polynomial to see if they yield zero.<br /><br />Let's start with \(x = 3\):<br />\[<br />12(3)^3 - 40(3)^2 - 27(3) + 90 = 108 - 360 - 81 + 90 = -243 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = -3\):<br />\[<br />3)^3 - 40(-3)^2 - 27(-3) + 90 = -108 - 360 + 81 + 90 = -297 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = 5\):<br />\[<br />12(5)^3 - 40(5)^227(5) + 90 = 3000 - 1000 - 135 + 90 = 1955 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = -5\):<br />\[<br />12(-5)^3 - 40(-5)^2 - 27(-5) + 90 = -3000 - 1000 + 135 + 90 = -3675 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = 2\):<br />\[<br />12(2)^3 - 40(2)^2 - 27(2) + 90 = 96 - 160 - 54 + 90 = -30 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = -2\):<br />\[<br />12(-2)^3 - 40(-2)^2 - 27(-2) + 90 = -96 - 160 + 54 + 90 = -192 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = 6\):<br />\[<br />12(6)^3 - 40(6)^2 - 27(6) + 90 = 2592 - 1440 - 162 + 90 = 960 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = -6\):<br />\[<br />12(-6)^3 - 40(-6)^2 - 27(-6) + 90 = -2592 - 1440 + 162 + 90 = -3880 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = 10\):<br />\[<br />12(10)^3 - 40(10)^2 - 27(10) + 90 = 12000 - 4000 - 270 + 90 = 7800 \quad (\text{not a root})<br />\]<br /><br />Next, let's try \(x = -10\):<br />\[<br />12(-10)^3 - 40(-10)^2