How would you completely factor: a(x) =8x^3 - 12x^2 + 18x -27 without knowing any of the roots?

Answer 1

We need to use a technique called factoring by grouping

Will write as 2 groups of terms 8x^3 -12x^2 and 18x -27 and factor each group separately

The first group has a common factor of 4x^2 so this factors to 4x^2( 2x -3)

The second group has a common factor of 9 so this factors to 9(2x -3) so now I have

(4x^2)(2x-3) +9(2x -3) Both groups have a common factor of (2x -3) so can rewrite as

(4x^2 +9)(2x -3) This is as far as we can factor. If we want to solve we set the equation equal to zero and get 4x^2 +9=0 2x -3 = 0

4x^2 = -9 2x =3

x^2 = -9/4 x=1.5

x= +/- (3/2)i

So have i real and 2 imaginary solutions. You can graph and confirm that there is only 1 zero

Answer 2

This can be very challenging depending on the problem but here is the process:

1) Use Descartes Rule of Signs to determine potential roots: There are 3 sign changes so there could be 3 positive real roots or 1 positive real root. Using a negative for "x" results in no sign changes therefore there will be no negative real roots.

2) Use the Rational Roots Theorem to find potential rational roots. They would be +/- the factors of 27 divided by the factors of 8. That means they could be +/- 3/8, 3/4, 1, 9/8, 3/2, 9/4, 3, 27/8, 9/2, 27/4, 9, 27/2, 27 but, since we already showed there are no negative real root, we can ignore all the negatives.

3) Start doing synthetic division on the possible rational roots. I chose to do 1 first. When I divide by 1 [really (x-1)], I find out it doesn't divide evenly. But I also find out the coefficients of the quotient are 8, -4, 14, -13 which follows a plus, minus, plus, minus pattern. That tells me there are no roots smaller than 1. The I tried 3. When I divide, I found it doesn't work either but I also find out the coefficients of the quotient are 8, 12, 54, 135 which are all positive so therefore there are no roots larger than 3. Next I chose 3/2 and it worked. So (x - 3/2) is a factor and the quotient is (8x² + 18).

4) Set 8x² + 18 equal to zero and solve to find the other roots are complex and are + 3/2i

Answer 3

With cubic equations, it's worth trying factoring by grouping.

a(x) = 8x³ - 12x² + 18x - 27

a(x) = (8x³ - 12x²) + (18x - 27)

Factor the greatest common factor out of each set of terms:

a(x) = 4x²(2x - 3) + 9(2x - 3)

a(x) = (4x² + 9)(2x-3)

If you are limited to the Real Number Set, you're done. In the Complex Number Set, however, (4x²+9) is a sum of squares and factors to (2x+3i)(2x-3i), so:

a(x) = (2x+3i)(2x-3i)(2x-3)

So a(x) has two complex roots and one real root.

Answer 4

If you want to factor a(x) = 8x³ - 12x² + 18x -27, there are a few different methods you can try. Since there are four terms in the equation, the first option to try is factoring by grouping.

a(x) = 8x³ - 12x² + 18x -27

First, we want to make two groups of two terms each

a(x) = (8x³ - 12x² ) + (18x -27 )

Next, we want to find the greatest common factor for each set of parentheses.

a(x) = 4x² (2x - 3) + 9 (2x - 3)

Factoring by grouping will ONLY work if the parentheses are now the same. Since they are, we're good to go! Now we can treat (2x - 3) as a factor that we can take out of each of our two terms. We can factor (2x - 3) out to the front, which will leave us with a (4x² + 9).

a(x) = (2x - 3) (4x² + 9)

Then we need to check if 4x² + 9 can be factored any more. It can't, so our final answer is:

a(x) = (2x - 3) (4x² + 9)

Answer 5

Well, as luck would have it...

(8x^3 - 12x^2) + (18x - 27)=

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

(4x^2 + 9) (2x - 3)