Mixed factors

Question

I have no idea how to do mixed factors: i have no idea how to approach these questions

Factorise:

3x^3 + 9x^2- 3x -9

 

x^4 - x^3 + 8x - 8

Answer 1

3x³ + 9x² - 3x - 9

3x²(x + 3) -3(x + 3)

(3x² - 3)(x + 3)

3(x² - 3)(x + 3)

 

Follow this to do the other.

Answer 2

3x^3 + 9x^2- 3x -9

 

First off I can immediately see that every term is divisible by 3.

Factor out a 3

 

3x^3 + 9x^2- 3x -9 = 3(x³ + 3x² - x - 3)

 

Have you ever used synthetic division?

The real roots of x³ + 3x² - x - 3 must be contained within the

factors of the constant.. -3 .. divided by the factors of the coefficient of the

term with largest exponent... 1

 

Possible real zeroes, or roots, are ±{-1/1, -3/1} = ±{-1, -3} ... {1, -1, 3, -3}

Do not forget the

So there are 4 possible real roots for x³ + 3x² - x - 3

 

Let's see if x=1 is a root.  If it is then (x-1) is a root

 

1  |  1    3    -1    -3      Your possible root divided into the coefficients

    |       1     4      3

    -----------------------

      1    4     3    | 0      Since the remainder is 0, x=1 is a root

                         ----

 

x³ - 3x² - x - 3 = (x-1)(x² + 4x + 3)

 

We can factor x² + 4x + 3

The factors of 3 that add to 4 are 3(1)

x² + 4x + 3 = (x+3)(x+1)

 

Putting this all together:

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

 

Can you try the second one?

 

x^4 - x^3 + 8x - 8

 

The possible real roots:  ±{(roots of  -8)/(roots of 1)}

Possible real roots:  ±{1/1, 2/1, 4/1, 8/1} = ±{1, 2, 4, 8}

So there are 8 total possible real roots to try.

 

Remember:  If x=1 is a root, then (x-1) is a factor

                   If x=-1 is a root, then (x+1) is a factor

                   And so forth