- P is the polynomial defined by P(x) =`-4+2*4*x^2-4*x^3`
- Compute P(1)
- Find the polynomial Q such that for any real x, P(x)=(x-1)Q(x)

This type of exercise can be solved using the function : factor

Algebraic calculus is a form of calculus that combines letters, numbers, and operations. The following formulas can be used in factoring and expanding algebraic expressions.

#### Formulas

- ka+kb=`k*(a+b)`
- ka-kb=`k*(a-b)`
#### Special expansions

- `(a+b)^2=a^2+2*a*b+b^2`
- `(a-b)^2=a^2-2*a*b+b^2`
- `(a-b)*(a+b)=a^2-b^2`
#### Factoring by (x-a)

Let a, b and k be three numbers we have :

Let a,b be two numbers we have the following three equalities: :

P is a polynomial, a a real. If P(a)=0, then P is factorizable by (x-a). That is, there exists a polynomial Q such that for all x, P(x)=(x-a)Q(x).

To simplify a literal expression, we group the dependent terms depending on the same letters and then we reduce each grouping, as in this example: x+x+3y-2y=2x+y.