Question 550815
<pre>
f(x) = 8x<sup>4</sup> - 24x<sup>3</sup> - 424x<sup>2</sup> - 72x

We set the right side = 0

8x<sup>4</sup> - 24x<sup>3</sup> - 424x<sup>2</sup> - 72x = 0

We can factor out 8x:

8x(x<sup>3</sup> - 3x<sup>2</sup> - 53x - 9) = 0

So by the zero factor principle:

8x = 0         x<sup>3</sup> - 3x<sup>2</sup> - 53x - 9 = 0
 x = 0

So we see that one rational solution is 0

Next we try to solve x<sup>3</sup> - 3x<sup>2</sup> - 53x - 9 = 0

If it has any rational solutions they must be
± some divisor of 9

Possible rational solutions of x<sup>3</sup> - 3x<sup>2</sup> - 53x - 9 = 0
are ±1, ±3, ±9

It has 1 sign change so we know it has one positive
solution.

Try x = 1

1 |1 -3 -53  -9
  |<u>   1  -2 -55</u>
   1 -2 -55 -64

No, 1 is not a solution, since the remainder is not 0

Try x = 3

3 |1 -3 -53   -9
  |<u>   3   0 -159</u>
   1  0 -53 -168

No, 3 is not a solution, since the remainder is not 0
 
Try x = 9

9 |1 -3 -53 -9
  |<u>   9  54  9</u>
   1  6   1  0

Hurray! 9 is a solution because the remainder is 0.

Now we know that 0 and 3 are rational zeros of f(x).

Now we have factored f(x) as

f(x) = 8x(x - 9)(x<sup>2</sup> + 6x + 1)

So all the rational zeros of f(x) are 0 and 9

However there are two irrational solutions obtainable by
setting x<sup>2</sup> + 6x + 1 = 0 and using the quadratic formula:

x = {{{(-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

x = {{{(-6 +- sqrt( 6^2-4*1*1 ))/(2*1) }}} 

x = {{{(-6 +- sqrt( 36-4 ))/2 }}}

x = {{{(-6 +- sqrt( 32 ))/2 }}}

x = {{{(-6 +- sqrt(16*2 ))/2 }}}

x = {{{(-6 +- 4sqrt(2))/2 }}}

x =  {{{(2(-3 +- 2sqrt(2)))/2 }}}

x =  {{{(cross(2)(-3 +- 2sqrt(2)))/cross(2) }}}

x =  {{{-3 +- 2sqrt(2)}}}

But you weren't asked for those two irrational zeros.

The only rational zeros are 0 and 9.

Edwin</pre>