# A-level Mathematics/OCR/C2/Dividing and Factoring Polynomials

< A-level Mathematics‎ | OCR‎ | C2

## Remainder TheoremEdit

The remainder theorem states that: If you have a polynomial f(x) divided by x + c, the remainder is equal to f(-c). Here is an example.

What will the remainder be if ${\displaystyle x^{3}+8x^{2}-4x^{2}+17x-40}$  is divided by x - 3?

$\displaystyle f(3)= 3^3 + 8 \left ( 3 \right )^2 - 4\left ( 3 \right )^2 + 17\left The remainder is 74. ==Factorising== When you factor an equation you try to "unmultiply" the equation. The N-Roots Theorem states that if f(x) is a polynomial of degree greater than or equal to 1, then f(x) has exactly n roots, providing that a root of multiplcity k is counted k times. The last part means that if an equation has 2 roots that are both 6, then we count 6 as 2 roots. ===The Factor Theorem=== The factor theorem allows us to check whether a number is a factor. It states: {{{{BOOKTEMPLATE}}/Remember|A polynomial [itex]f(x)$ has a factor x - c if and only if ${\displaystyle f(c)=0}$ .}}

For example:

Determine if x + 2 is a factor of ${\displaystyle 2x^{2}+3x-2}$ .

Since c is positive instead of negative we need to use this basic identity:

${\displaystyle x+2=x-\left(-2\right)}$

Now we can use the factor theorem.

${\displaystyle 2\left(-2\right)^{2}+3\left(-2\right)-2=8-6-2=0}$ .

Since the resultant is 0, (x+2) is a factor of ${\displaystyle 2x^{2}+3x-2}$ .

This means it is possible to re-state the polynomial in the form (x+2)( some linear expression of x).

So ${\displaystyle 2x^{2}+3x-2}$  = (x+2)(ax+b)

Expanding the right hand side we get :

${\displaystyle 2x^{2}+3x-2}$  = ${\displaystyle ax^{2}+x(2a+b)+2b}$

Equating like terms we get :

2= a

2a+b = 3 and

2b = -2

Giving a= 2, b= -1 from the first and third equations and this works in the second, so

${\displaystyle 2x^{2}+3x-2}$  = (x+2)(2x-1)

This is part of the C2 (Core Mathematics 2) module of the A-level Mathematics text. Appendix A: Formulae