what is the factor of this polynomial?

Question

g(x)= x^4 - 4x^3 - 2x^2 +12x +9
p(x)= x^5+2x^4+x^3-x^2-2x-1

Answer 1

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

By trial and error: (x+1) is factor

g(x) = (x + 1)(x^3 - 5x^2 + 3x + 9)

Again (x+1) is a factor of x^3 - 5x^2 + 3x + 9

g(x) = (x + 1)(x + 1)(x^2 - 6x + 9)

g(x) = (x + 1)(x + 1)(x - 3)(x - 3)

You can solve the second one in a similar fashion.

Answer 2

1) g(x) = x^4 - 4x^3 - 2x^2 + 12x + 9

If there are factors, they will be of the form (x + a),
where 'a' may be any of the factors of 9.
All the factors of 9 are -1, +1, -3, +3, -9 and +9.
So try each one in turn as a value for x
and see if g(x) becomes zero.

x = -1 implies g(-1) = (-1)^4 - 4(-1)^3 - 2(-1)^2 + 12(-1) + 9
Therefore, g(-1) = 0. So if x = -1, then a = +1.
This means that (x + 1) is a factor.

Now divide (x + 1) into g(x) by synthetic division.
The answer becomes : x^3 - 5x^2 + 3x + 9.

Now before going onto the next possible factor,
it's necessary to try the same one again,
because (x + 1)^2 may be a factor of g(x).

x = -1 implies x^3 - 5x^2 + 3x + 9 = (-1)^3 - 5(-1)^2 + 3(-1) + 9
Again, this value is zero, so another (x + 1) is a factor.

Dividing (x^3 - 5x^2 + 3x + 9) by (x + 1) gives x^2 - 6x + 9.

Again substituting x = -1 into the last expression
gives a value of 16, which is no longer zero, so that
exhausts the possibility of any more (x + 1) factors.

You could try the others, but I think it's at a
point where you should be able to see that
x^2 - 6x + 9 factors as (x - 3)^2.

Thus, g(x) = (x + 1)^2 * (x - 3)^2
------------------------------------------------------------------------------------------
2) p(x) = x^5 + 2x^4 + x^3 - x^2 - 2x - 1

This is a lot easier to start with, because
we only have to try (x + 1) and (x - 1).

If x = -1, then p(-1) = (-1)^5 + 2(-1)^4 + (-1)^3 - (-1)^2 - 2(-1) - 1
So, p(-1) = 0, so (x + 1) is a factor.

Dividing (x + 1) into p(x) gives : x^4 + x^3 - x - 1

Trying again with x = -1 gives another value of zero.
Thus (x + 1) is again a factor,
so (x + 1)^2 is now a factor of p(x).

Dividing out again gives : x^3 - 1

This should be recognised as a difference of two cubes,
the factorisation being : (x - 1)(x^2 + x + 1).

Thus, p(x) = (x - 1)(x + 1)^2 * (x^2 + x + 1)

You could go further and factorise (x^2 + x + 1),
by letting it equal zero and applying the quadratic formula.
x^2 + x + 1 = 0
So, x = {-1 ± √[1 - 4(1)(1)]} / 2
= [-1 ± √(-3)] / 2
= -1/2 ± i*√(3) / 2
which gives :
p(x) = (x - 1)(x + 1)^2 * [x + 1/2 + i√(3) / 2] [x + 1/2 - i√(3) / 2]