List all possible rational zeros given by the rational zeros theorem

Solution:

We can use the rational zero theorem to find rational zeros of a polynomial.

By Rational Zero Theorem,

If P(x) is a polynomial with integer coefficients and if is a zero of P(x) (P( ) = 0), 

then p is a factor of the constant term of P(x)

q is a factor of the leading coefficient of P(x)

Possible value of rational zero is p/q

Given, f(x) = 2x3 + 8x2 + 7x - 8

Here, constant term, p = -8

Leading coefficient, q = +2

The factors of the constant term -8 are ±1, ±2, ±4, ±8.

The factor of the leading coefficient is ±1 and ±2.

Possible values of rational zeros p/q = ±1/±1, ±2/±1, ±4/±1, ±8/±1, ±1/±2, ±4/±2, ±8/±2. 

Therefore, the values of possible rational zeros are ±1, ±2, ±4, ±8.

Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function. f(x) = 2x3 +8x2+7x-8

Summary:

Using the rational zeros theorem, all possible rational zeros of the function f(x) = 2x3 + 8x2 + 7x - 8 are ±1, ±2, ±4, ±8.

The way to determine all possible roots is through The Rational Roots Theorem, which states:

If P(x) is a polynomial with integer coefficients and if #p/q# is a zero of #P(x) ( P(p/q) = 0 )#, then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient.

Basically, find all integer (positive AND negative) factors of both the constant term (6, in this case) and the leading coefficient (2, in this case), and find all possible quotients #p/q# where p is a factor of the constant, and q is a factor of the leading coefficient.

The integer factors of 6 are: #+-1, +-2, +-3, +-6#
The integer factors of 2 are: #+-1, +-2#

Therefore, the possible rational roots are:
#+-1, +-2, +-3, +-6, +-1/2, +-3/2#

To find which of these are roots in the actual equation, you could use Guess and Check with Synthetic Division, or simply group and factor the polynomial. I'll show the latter for the sake of space preservation:

#2x^3 - 4x - 3x^2 + 6#
# = 2x(x^2 - 2) - 3(x^2 - 2)#
# = (2x - 3)(x^2 - 2)#
From this we can conclude that one rational root is #3/2#, which is one of the roots we foretold!

However, we are still stuck with #(x^2 - 2)#. Fortunately, we can use a difference of squares formula to factor this into two linear factors:

#(x - sqrt2)(x + sqrt2)#

So your complete factorization of this polynomial is:

#f(x) = (2x - 3)(x - sqrt2)(x + sqrt2)

From this, we can determine that the roots of this equation are #3/2 and +-sqrt2#.

Given:

#f(x) = x^4+2x^3-12x^2-40x-32#

Rational roots theorem

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-32# and #q# a divisor of the coefficient #1# of the leading term.

That means the the only possible rational zeros are:

#+-1, +-2, +-4, +-8, +-16, +-32#

Descartes' Rule of Signs

The pattern of signs of the coefficients of #f(x)# is #+ + - - -#. With one change of signs, Descartes' Rule of Signs tells us that #f(x)# has exactly one positive real zero.

The pattern of signs of coefficients of #f(-x)# is #+ - - + -#. With #3# changes of sign, Descartes' Rule of Signs allows us to deduce that #f(x)# has #3# or #1# negative real zero.

Bonus - Find the actual zeros

Note that the coefficient of #x^4# is odd, but all the other coefficients are even. Therefore any integer zero of #f(x)# must be even.

We find:

#f(-2) = (-2)^4+2(-2)^3-12(-2)^2-40(-2)-32 = 16-16-48+80-32 = 0#

So #-2# is a zero and #(x+2)# a factor:

#x^4+2x^3-12x^2-40x-32 = (x+2)(x^3-12x-16)#

We find that #-2# is also a zero of the remaining cubic expression:

#(-2)^3-12(-2)-16 = -8+24-16 = 0#

So #(x+2)# is a factor again:

#x^3-12x-16 = (x+2)(x^2-2x-8)#

Finally, to factor the remaining quadratic, note that #4*2 = 8# and #4-2=2#, so we find:

#x^2-2x-8 = (x-4)(x+2)#

So:

#f(x) = (x+2)^3(x-4)#

has zeros #-2# with multiplicity #3# and #4# with multiplicity #1#.

Roots of a Polynomial

A root or zero of a function is a number that, when plugged in for the variable, makes the function equal to zero. Thus, the roots of a polynomial P(x) are values of x such that P(x) = 0.

The Rational Zeros Theorem

The Rational Zeros Theorem states:

If P(x) is a polynomial with integer coefficients and if

is a zero of P(x) (P(

) = 0), then p is a factor of the constant term of P(x) and q is a factor of the leading coefficient of P(x).

We can use the Rational Zeros Theorem to find all the rational zeros of a polynomial. Here are the steps:

1. Arrange the polynomial in descending order
2. Write down all the factors of the constant term. These are all the possible values of p.
3. Write down all the factors of the leading coefficient. These are all the possible values of q.
4. Write down all the possible values of
   
   . Remember that since factors can be negative,
   
   and -
   
   must both be included. Simplify each value and cross out any duplicates.
5. Use synthetic division to determine the values of
   
   for which P(
   
   ) = 0. These are all the rational roots of P(x).

Example: Find all the rational zeros of P(x) = x3 -9x + 9 + 2x4 -19x2.

1. P(x) = 2x4 + x3 -19x2 - 9x + 9
2. Factors of constant term: ±1, ±3, ±9.
3. Factors of leading coefficient: ±1, ±2.
4. Possible values of
   
   : ±
   
   , ±
   
   , ±
   
   , ±
   
   , ±
   
   , ±
   
   . These can be simplified to: ±1, ±
   
   , ±3, ±
   
   , ±9, ±
   
   .
5. Use synthetic division:
   

Thus, the rational roots of P(x) are x = - 3, -1,

We can often use the rational zeros theorem to factor a polynomial. Using synthetic division, we can find one real root a and we can find the quotient when P(x) is divided by x - a. Next, we can use synthetic division to find one factor of the quotient. We can continue this process until the polynomial has been completely factored.

Example (as above): Factor P(x) = 2x4 + x3 -19x2 - 9x + 9.

As seen from the second synthetic division above, 2x4 + x3 -19x2 -9x + 9÷x + 1 = 2x3 - x2 - 18x + 9. Thus, P(x) = (x + 1)(2x3 - x2 - 18x + 9). The second term can be divided synthetically by x + 3 to yield 2x2 - 7x + 3. Thus, P(x) = (x + 1)(x + 3)(2x2 - 7x + 3). The trinomial can then be factored into (x - 3)(2x - 1). Thus, P(x) = (x + 1)(x + 3)(x - 3)(2x - 1). We can see that this solution is correct because the four rational roots found above are zeros of our result.

What is the rational zero theorem formula?

What does possible rational zeros mean?