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18 January, 18:40

For the function y=x^5+1x^3-30x, find all real zeros.

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  1. 18 January, 20:46
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    The real zeroes are - √5, 0, √5

    Step-by-step 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
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