Answer:
The real zeroes are -√5 , 0 , √5
Explanation:
* Lets explain how to solve the problem
- The function is y = x^5 + x³ - 30x
- Zeros of any equation is the values of x when y = 0
- To find the zeroes of the function equate y by zero
∴ x^5 + x³ - 30x = 0
- To solve this equation factorize it
∵ x^5 + x³ - 30x = 0
- There is a common factor x in all the terms of the equation
- Take x as a common factor from each term and divide the terms by x
∴ x(x^5/x + x³/x - 30x/x) = 0
∴ x(x^4 + x² - 30) = 0
- Equate x by 0 and (x^4 + x² - 30) by 0
∴ x = 0
∴ (x^4 + x² - 30) = 0
* Now lets factorize (x^4 + x² - 30)
- Let x² = h and x^4 = h² and replace x by h in the equation
∴ (x^4 + x² - 30) = (h² + h - 30)
∵ (x^4 + x² - 30) = 0
∴ (h² + h - 30) = 0
- Factorize the trinomial into two brackets
- In trinomial h² + h - 30, the last term is negative then the brackets
have different signs ( + )( - )
∵ h² = h × h ⇒ the 1st terms in the two brackets
∵ 30 = 5 × 6 ⇒ the second terms of the brackets
∵ h × 6 = 6h
∵ h × 5 = 5h
∵ 6h - 5h = h ⇒ the middle term in the trinomial, then 6 will be with
(+ ve) and 5 will be with (- ve)
∴ h² + h - 30 = (h + 6)(h - 5)
- Lets find the values of h
∵ h² + h - 30 = 0
∴ (h + 6)(h - 5) = 0
∵ h + 6 = 0 ⇒ subtract 6 from both sides
∴ h = -6
∵ h - 5 = 0 ⇒ add 5 to both sides
∴ h = 5
* Lets replace h by x
∵ h = x²
∴ x² = -6 and x² = 5
∵ x² = -6 has no value (no square root for negative values)
∵ x² = 5 ⇒ take √ for both sides
∴ x = ± √5
- There are three values of x ⇒ x = 0 , x = √5 , x = -√5
∴ The real zeroes are -√5 , 0 , √5