# The possible rational zeros of P(x) = 14x3 + 5x2 − 19x − 26 are?

### 5 Answers

- la consoleLv 75 months ago
14x³ + 5x² - 19x - 26 = 0 → it's necessary to eliminate the term at the power 2

14x³ + 5x² - 19x - 26 = 0 → let: x = z - (5/42)

14.[z - (5/42)]³ + 5.[z - (5/42)]² - 19.[z - (5/42)] - 26 = 0

14.[z - (5/42)]².[z - (5/42)] + 5.[z² - (5/21).z + (25/1764)] - 19z + (95/42) - 26 = 0

14.[z² - (5/21).z + (25/1764)].[z - (5/42)] + 5z² - (25/21).z + (125/1764) - 19z + (95/42) - 26 = 0

14.[z³ - (5/42).z² - (5/21).z² + (25/882).z + (25/1764).z - (125/74088)] + 5z² - (25/21).z + (125/1764) - 19z + (95/42) - 26 = 0

14.[z³ - (5/14).z² + (25/588).z - (125/74088)] + 5z² - (25/21).z + (125/1764) - 19z + (95/42) - 26 = 0

14z³ - 5z² + (25/42).z - (125/5292) + 5z² - (25/21).z + (125/1764) - 19z + (95/42) - 26 = 0

14z³ - (823/42).z - (125372/5292) = 0 ← no term with power 2

14z³ - (823/42).z - (125372/5292) = 0 → let: z = u + v

14.(u + v)³ - (823/42).(u + v) - (125372/5292) = 0

14.(u + v)².(u + v) - (823/42).(u + v) - (125372/5292) = 0

14.(u² + 2uv + v²).(u + v) - (823/42).(u + v) - (125372/5292) = 0

14.[u³ + u²v + 2u²v + 2uv² + uv² + v³] - (823/42).(u + v) - (125372/5292) = 0

14.[u³ + v³ + 3u²v + 3uv²] - (823/42).(u + v) - (125372/5292) = 0

14.[(u³ + v³) + (3u²v + 3uv²)] - (823/42).(u + v) - (125372/5292) = 0

14.[(u³ + v³) + 3uv.(u + v)] - (823/42).(u + v) - (125372/5292) = 0

14.(u³ + v³) + 42uv.(u + v) - (823/42).(u + v) - (125372/5292) = 0 → you can factorize: (u + v)

14.(u³ + v³) + (u + v).[42uv - (823/42)] - (125372/5292) = 0 → suppose that: [42uv - (823/42)] = 0 ← equation (1)

14.(u³ + v³) + (u + v).[0] - (125372/5292) = 0

14.(u³ + v³) - (125372/5292) = 0 ← equation (2)

You obtain a system of 2 equations:

(1) : [42uv - (823/42)] = 0

(1) : 42uv = 823/42

(1) : uv = 823/1764

(1) : u³v³ = (823/1764)³

(2) : 14.(u³ + v³) - (125372/5292) = 0

(2) : 14.(u³ + v³) = 125372/5292

(2) : u³ + v³ = 125372/74088

(2) : u³ + v³ = 31343/18522

Let: U = u³

Let: V = v³

You obtain a new system of 2 equations:

(1) : UV = (823/1764)³ ← this is the product P

(2) : U + V = 31343/18522 ← this is the sum S

You know that the values U & V are the solutions of the following equation:

x² - Sx + P = 0 ← don’t confuse with the item x (initial equation)

x² - (31343/18522).x + (823/1764)³ = 0

Δ = (31343/18522)² - [4 * (823/1764)³]

Δ = (31343²/18522²) - [(4 * 823³)/1764³]

Δ = [(16 * 31343²) - (4 * 823³)]/1764³ → you know that: 1764 = 42² → 1764³ = 42⁶

Δ = 13488371316/42⁶

Δ = (42² * 7646469)/42⁶

Δ = 7646469/42⁴

x₁ = [(31343/18522) + (1/42²).√7646469]/2

x₁ = (31343/37044) + (1/3528).√7646469 ← this is U → recall: U = u³ → u = U^(1/3)

u = [(31343/37044) + (1/3528).√7646469]^(1/3)

x₂ = [(31343/18522) - (1/42²).√7646469]/2 ← this is V → recall: V = v³ → v = V^(1/3)

v = [(31343/37044) - (1/3528).√7646469]^(1/3)

Recall: z = u + v

Recall: x = z - (5/42)

x = u + v - (5/42)

x = [(31343/37044) + (1/3528).√7646469]^(1/3) + [(31343/37044) - (1/3528).√7646469]^(1/3) - (5/42)

x ≈ 1.1768468560802 + 0.396443501184035 - (5/42)

x ≈ 1.45424273821662

- sepiaLv 75 months ago
P(x) = 14x3 + 5x2 − 19x − 26

P(x) = (x - 1) x (14 x + 19) - 26

Real root:

x ≈ 1.4542

Complex roots:

x ≈ -0.90569 - 0.67585 i

x ≈ -0.90569 + 0.67585 i

- DixonLv 75 months ago
There is only one real root, it's not rational and it is a mess (so are the complex roots). I suggest you have copied it wrong.

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- Anonymous5 months ago
Are you seriously that lazy?

Thank you so much! didn't know where I was going wrong