List all the possible rational zeros of p(x)=2x3-3x2+4x-6?

Question

Problem is p(x)= 2x(cubed) - 3x(squared) + 4x - 6

Choices for answer:

1) +/-1, +/-2, +/-3, +/-6, +/- 1/2, +/- 3/2

2) +/-1, +/-2, +/-6, +/- 1/3, +/- 1/2, +/- 2/3

3) +/-1, +/-2, +/-3, +/-6

4) +/-1, +/-2, +/-6, +/- 1/3, +/- 1/2, +/- 2/3

Answer 1

You should not have plugged in the values into equation to tell NO! just look, they suggest you 4 sets of numbers with more than 3 numbers in a row, while your equation has power 3 which means that only 3 solutions are really possible!

Answer 2

2x³ - 3x² + 4x - 6 = 0 ? permit: x = z + (a million/2) 2[z + (a million/2)]³ - 3[z + (a million/2)]² + 4[z + (a million/2)] - 6 = 0 2{ [z + (a million/2)]² * [z + (a million/2)] } - 3[z² + z + (a million/4)] + 4z + 2 - 6 = 0 2{ [z² + z + (a million/4)] * [z + (a million/2)] } - 3z² - 3z - (3/4) + 4z - 4 = 0 2{ z³ + (z²/2) + z² + (z/2) + (z/4) + (a million/8) } - 3z² + z - (19/4) = 0 2{ z³ + (3z²/2) + (3z/4) + (a million/8) } - 3z² + z - (19/4) = 0 2z³ + 3z² + (3z/2) + (a million/4) - 3z² + z - (19/4) = 0 2z³ + (5z/2) - (9/2) = 0 ? you multiply by 2 4z³ + 5z - 9 = 0 ? permit: z = (u + v) 4(u + v)³ + 5(u + v) - 9 = 0 4[(u + v)²(u + v)] + 5(u + v) - 9 = 0 4[(u² + 2uv + v²)(u + v)] + 5(u + v) - 9 = 0 4[u³ + u²v + 2u²v + 2uv² + uv² + v³] + 5(u + v) - 9 = 0 4[u³ + v³ + 3u²v + 3uv²] + 5(u + v) - 9 = 0 4[u³ + v³ + 3uv(u + v)] + 5(u + v) - 9 = 0 4(u³ + v³) + 12uv(u + v) + 5(u + v) - 9 = 0 4(u³ + v³) + (u + v)(12uv + 5) - 9 = 0 ? permit: (12uv + 5) = 0 ? equation (a million) 4(u³ + v³) - 9 = 0 ? equation (2) You get a device of two equatinos: (a million) : 12uv + 5 = 0 (a million) : uv = - 5/12 (a million) : u³v³ = - (5/12)³ (2) : 4(u³ + v³) - 9 = 0 (2) : u³ + v³ = 9/4 ? permit: U = u³ ? permit: V = v³ (a million) : UV = - (5/12)³ (2) : U + V = 9/4 you realize the product P and the sum S of two numbers. those numbers are the answer of the equation: x² - Sx + P = 0 x² - (9/4)x - (5/12)³ = 0 ? = (- 9/4)² - 4[a million * - (5/12)³] ? = (80 one/sixteen) + [4 * (5³/12³)] ? = (80 one/sixteen) + [4 * (one hundred twenty five/1728)] ? = (80 one/sixteen) + (one hundred twenty five/432) ? = 2312/432 = 289/fifty 4 = (289/9) * (a million/6) ? = (17/3)² * (a million/6) = [17/(3?6)]² = [(17?6)/18]² x1 = [ (9/4) - {(17?6)/18} ] / 2 = [ (80 one/36) - {(34?6)/36} ] / 2 = (80 one - 34?6)/seventy two x2 = [ (9/4) + {(17?6)/18} ] / 2 = [ (80 one/36) + {(34?6)/36} ] / 2 = (80 one + 34?6)/seventy two U = (80 one - 34?6)/seventy two ? u = ³?U ? u = ³?[(80 one - 34?6)/seventy two] V = (80 one + 34?6)/seventy two ? v = ³?V ? v = ³?[(80 one + 34?6)/seventy two] bear in mind: z = u + v z = ³?[(80 one - 34?6)/seventy two] + ³?[(80 one + 34?6)/seventy two] z = a million bear in mind: x = z + (a million/2) x = a million + (a million/2) ? x = 3/2 ? P(3/2) = 0

Answer 3

I think, something is wrong in the question.
P(+1)= 2*1 - 3*1 + 4*1 - 6 = -3
That means P(1)=-3 and P(1) is not equal to zero.
So, all of the alternatives are wrong...

Answer 4

None of the above. Try +1. It gives a value of -3. Therefore, none of those answers fits. Try again.