F(x)= 2x^3 + 9x^2 + x-12

Answer 1

Here is the ENTIRE story from the beginning...

There are 2 sets

P = {factors of 12} = { +-1, +-2, +-3, +-4, +-6, +-12}

Q = {factors of 2} = {+-1, +-2}

P/Q = {+-1, +-2, +-3, +-4, +-6, +-12, +-1/2, +-3/2, +-3/4}

rational roots theorem says that these are the ONLY possible rational roots.

Each one (both positive and negative) must be tested, one at a time, to see

which one makes the polynomial equal to zero. Yes, as many as all 18 must

be checked unless you find one that works.

However, as Mark suggested, -4 is one that does make the polynomial zero, because

2(-4)^3 + 9(-4)^2 + -4 - 12 =

2(-64) + 9(16) + -4 - 12 =

-128 + 144 + -4 - 12 =

16 + -4 - 12 =

12 - 12 =

0

So x=-4 is a solution which means x+4 is a factor

Again, as Mark suggested, synthetic division says:

-4 | 2 9 1 -12

-8 -4 12

-----------------------

2 1 -3 0

so the resulting quadratic is 2x^2 + x - 3 = 0

which factors ( 2x + 3)(x - 1 ) = 0

the other two solutions are x = -3/2 and x = 1

notice they appear in the list of P/Q....

so yes we could have started with x=1

thanks Mark