```Factorising Cubics, the remainder theorem, the factor theorem
and CAS
1. Consider the polynomial p( x)  3x 3  x  3 . We will ask our calculator to remember
this as p (x ) .
Using the active menu select command, define
Now evaluate
p ( 2)
p ( 3)
p ( 4)
1
p( )
2
2. Type in the fraction as shown—
Now highlight the fraction and go to the
propFrac.
What does our answer tell us?
Repeat for
p( x)
.
( x  3)
3. Now consider the polynomial p( x)  x 3  2 x 2  5x  6
This time we want to factorise the polynomial.
This can be achieved two ways—
Method 1 using interactive, transformation and factor—
Method 2—define p( x)  x 3  2 x 2  5x  6 then solve for p ( x)  0 using interactive,
This gives solutions of x  2, x  1, x  3
meaning the polynomial has factors of
( x  2) ( x  1) ( x  3)
Try both methods for x 3  4 x 2  x  6 .
4. Consider the question—
When P( x)  x 3  2 x  a is divided by x  2 the remainder is 4. Find the value of a.
This can be done using CAS.
The remainder theorem tells us that we need to set up p (2)  4 and solve for a.
We define the polynomial (active, command, define) then enter p (2)  4 , highlight,
interactive, advanced solve BUT we need to solve for a.
```