# List all possible rational zeros for the polynomial, find all real zeros and factor completely.  f(x)=5x⁴+23x³+13x²-23x-18PLEASE SHOW ALL WORK!!!!!!

Asked on by nanalisa

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to perform the rational roots test, hence you need to form a set of possible rational roots of polynomial such that: possible rational root = factors of constant term/factors of leading coefficient.

Since the constant term is 18, hence you need to list the factors of 18 such that:  `{+-1;+-2+-3;+-6;+-9;+-18}` .

You  need to list the factors of leading coefficient 5 such that: `{+-1;+-5}.`

Hence, you may find the rational roots among `{+-1/5;+-2/5;+-3/5;+-6/5;+-9/5;+-18/5}`

You need to substitute each of these fractions in polynomial to check if it cancels.

Considering `x = 1/5`  yields:

`f(1/5)= 5*(1/5)^4 + 23*(1/5)^3 + 13*(1/5)^2 - 23/5 - 18`

`f(1/5)= (5 + 115 + 325 - 2875-11250)/625 !=0`

Considering`x = -1/5`  yields:

`f(-1/5)= 5*(-1/5)^4 + 23*(-1/5)^3 + 13*(-1/5)^2+ 23/5 - 18`

`f(-1/5)= (5 - 115 + 325 + 2875 - 11250)/625!=0`

You need to try`x = -2/5`  such that:

`f(-2/5)= 5*(-2/5)^4 + 23*(-2/5)^3 + 13*(-2/5)^2+ 23*2/5 - 18`

`` `f(-2/5)= (80 - 920 + 1300 + 5750 - 11250)/625 != 0`

`f(2/5)= (80+ 920 + 1300- 5750 - 11250)/625 != 0`

You need to try `x = -3/5`  such that:

`f(-3/5)= 5*(-3/5)^4 + 23*(-3/5)^3 + 13*(-3/5)^2+ 23*3/5 - 18`

`` `f(-3/5)= (405 - 621 + 2925 + 8625 - 11250)/625 != 0`

`f(3/5)= (405 +621 + 2925- 8625 - 11250)/625 != 0`

You need to try `x = -6/5`  such that:

`f(-6/5)= 5*(-6/5)^4 + 23*(-6/5)^3 + 13*(-6/5)^2+ 23*6/5 - 18`

`f(-6/5)= (6480 -4968 +11700 + 17250 - 11250)/625 != 0`

`f(6/5)= (6480+4968 +11700- 17250 - 11250)/625 != 0`

You need to try`x = -9/5`  such that:

`f(-9/5)= 5*(-9/5)^4 + 23*(-9/5)^3 + 13*(-9/5)^2+ 23*9/5 - 18`

`` `f(-9/5)= (32805 -83835 + 2925 + 25875 - 11250)/625 != 0`

`f(-9/5)= (32805 +83835 + 2925- 25875 - 11250)/625 != 0`

You need to try `x = -18/5`  such that:

`f(-18/5)= 5*(-18/5)^4 + 23*(-18/5)^3 + 13*(-18/5)^2+ 23*18/5 - 18 != 0`

`f(18/5)= 5*(18/5)^4 + 23*(18/5)^3 + 13*(18/5)^2+ 23*18/5 - 18 != 0`

You need to try x = -2 such that:

`f(-2)= 5*(-2)^4 + 23*(-2)^3 + 13*(-2)^2+ 23*2 - 18`

`f(-2)= 80 - 184 + 52 + 46 - 18 != 0`

You need to try x = 2 such that:

`f(2)= 80+ 184 + 52- 46 - 18 != 0`

You need to try x = -3 such that:

`f(-3)= 5*(-3)^4 + 23*(-3)^3 + 13*(-3)^2+ 23*3 - 18`

`f(-3)= 405 - 621 + 117 + 69 - 18 != 0`

You need to try x = -6 such that:

f`(-6)= 5*(-6)^4 + 23*(-6)^3 + 13*(-6)^2+ 23*6 - 18`

`f(-6)= 6480 - 4968 + 468 + 138 - 18!= 0`

Notice that the fractions `{+-1/5;+-2/5;+-3/5;+-6/5;+-9/5;+-18/5}`  fail to check the polynomial, hence the polynomial has no rational roots.

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