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

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

if you use photomath right away it will give you the answer not like this app that you have to wait for people.

-2

7 - 9 = -2

reversed:

-2 + 9 = 7