List all possible rational zeros for f(x) = 2x^4 - 21x^3 + 69x^2 - 56x - 48. Also, factor completely.  

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embizze's profile pic

embizze | High School Teacher | (Level 1) Educator Emeritus

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The answer above has the correct factorization and solutions. You also asked for all possible rational roots.

The rational root theorem states that for any root of the form `p/q` with `p,q` both integers, `p` is a factor of the constant term and `q` is a factor of the leading coefficient. (We assume the polynomial has real coefficients and is written in standard form).

Thus the possible rational roots have numerators that are factors of 48, and denominators that are factors of 2. The list:

`+-1,+-2,+-3,+-4,+-6,+-8,+-12,+-16,+-24,+-48,+-1/2,+-3/2`

* Note that it is unnecessary to list such possible roots as `2/2` as 1 is already included. *

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The given polynomial is 2x^4 - 21x^3 + 69x^2 - 56x - 48

2x^4 - 21x^3 + 69x^2 - 56x - 48

=> 2x^4 - 6x^3 - 15x^3 + 45x^2 + 24x^2 - 72x + 16x - 48

=> 2x^3(x - 3) - 15x^2(x - 3) + 24x(x - 3) + 16(x - 3)

=> (x - 3)(2x^3 - 15x^2 + 24x + 16)

=> (x- 3)(2x^3 + x^2 - 16x^2 - 8x + 32x + 16)

=> (x - 3)(x^2(2x + 1) - 8x(2x + 1) + 16(2x + 1))

=> (x - 3)(2x + 1)(x^2 - 8x + 16)

=> (x - 3)(2x + 1)(x - 4)^2

Equating this to 0 gives the roots x = 3, x = -1/2 and x = 4

The factorized form of 2x^4 - 21x^3 + 69x^2 - 56x - 48 is (x - 3)(2x + 1)(x - 4)^2 and the roots are x = -1/2, x = 3 and x = 4

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