function f(x)=6x^4+x^3-x^2+75x-25 has 4 distinct zeros. 2 0f the zeros are x = -5/2 and x = 1/3. What are the 2 other zeros?

Hello,

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► Solving quartic equations like yours is hard. When roots are provided you ought to use the factor theorem.

The factor theorem states that a polynomial 𝑓(𝑥) has a factor (𝑥–𝑘) if and only if 𝑓(𝑘)=0 (i.e. 𝑘 is a zero).

In your case, you have a function 𝑓(𝑥) defined by:

   𝑓(𝑥) = 6𝑥⁴ + 𝑥³ – 𝑥² + 75𝑥 – 25

You are told it has -5/2 and 1/3 as zeros.

Using the theorem, you can thus state that 𝑓(𝑥) has (𝑥+5/2) and (𝑥–⅓) as factors.

► Thus you need to use any method (e.g.polynomial long division) to factor out the factors you know exist.

   𝑓(𝑥) = 6𝑥⁴ + 𝑥³ – 𝑥² + 75𝑥 – 25

      = 6𝑥⁴ – 2𝑥³ + 3𝑥³ – 𝑥² + 75𝑥 – 25

      = 2𝑥³(3𝑥 – 1) + 𝑥²(3𝑥 – 1) + 25(3𝑥 – 1)

      = (3𝑥 – 1)(2𝑥³ + 𝑥² + 25)

      = (3𝑥 – 1)(2𝑥³ + 5𝑥² – 4𝑥² + 10𝑥 – 10𝑥 + 25)

      = (3𝑥 – 1)[𝑥²(2𝑥 + 5) – 2𝑥(2𝑥 + 5) – 5(2𝑥 + 5)]

      = (3𝑥 – 1)(2𝑥 + 5)(𝑥² – 2𝑥 – 5)

► Now you are left with a quadratic equation which you can solve using any method of your choice (e.g. quadratic formula, complete the square)

   𝑓(𝑥) = (3𝑥 – 1)(2𝑥 + 5)(𝑥² – 2𝑥 – 5)

      = (3𝑥 – 1)(2𝑥 + 5)(𝑥² – 2𝑥 + 1 – 6)             ← Complete the square

      = (3𝑥 – 1)(2𝑥 + 5)[(𝑥 – 1)² – 6]                   ← Since 𝑎²–2𝑎𝑏+𝑏²=(𝑎–𝑏)²

      = (3𝑥 – 1)(2𝑥 + 5)[(𝑥 – 1)² – (√6)²]             ← Express remainder as a square

      = (3𝑥 – 1)(2𝑥 + 5)(𝑥 – 1 – √6)(𝑥 – 1 + √6) ← Since 𝑎²–𝑏²=(𝑎–𝑏)(𝑎+𝑏)

► Then finding the roots of 𝑓(𝑥) is finding the values such that 𝑓(𝑥)=0:

   𝑓(𝑥) = 0

   6𝑥⁴ + 𝑥³ – 𝑥² + 75𝑥 – 25 = 0

   (3𝑥 – 1)(2𝑥 + 5)(𝑥 – 1 – √6)(𝑥 – 1 + √6) = 0

Resulting in a product of factors that is nil.

This allows us to use zero product property:

Any product that is nil means that at least one of its factors is nil.

Thus we must have:

            3𝑥 – 1 = 0

   OR   2𝑥 + 5 = 0

   OR   𝑥 – 1 – √6 = 0

   OR   𝑥 – 1 + √6 = 0

Solving those yields:

           𝑥 = ⅓

   OR   𝑥 = -⁵/₂

   OR   𝑥 = 1 + √6

   OR   𝑥 = 1 – √6

► Since any of those values will make 𝑓(𝑥)=0, they are all zeros of 𝑓(𝑥).

And since you already know that ⅓ and -⁵/₂ are zeros of 𝑓(𝑥), you conclude the two remaining zeros are:

   1–√6   and    1+√6

And visibly, those zeros are neither imaginary nor rational: they are irrational.

Explicatively,

Dragon.Jade :-)