The Factor Theorem Flashcards
1
Q
Content for factor theorem:
A
- Given polynomial f ( k ), and a constant k
- If f ( k ) = 0, then ( x - k ) is the factor of f ( k )
- ( If f ( 1 ) = 0, then ( x - 1 ) is the factor of f ( 1 ) )
- If ( x - k ) is a factor of f ( k ), then f ( k ) = 0
- ( If ( x - 1 ) is a factor of f ( k ), then f ( 1 ) = 0 )
Steps:
- Substitute values into the polynomial until you find f ( k ) = 0
- Divide the polynomial by ( x - k ) to find the quadratic
- Write f ( k ) = ( x - k ) ( ax^2 + bx + c )
- Factorise the quadratic to have it in linear factors ( if possible )
2
Q
Show that ( x - 2 ) is a factor of p( x ) = x^3 + x^2 - 4x - 4. Hence factorise p( x ) fully.
A
- ( x - 2 ) so x = 2
- ( Substitute x = 2 into equation and it should = 0 )
- x - 2 | x^3 + x^2 - 4x - 4 | ( Dividing )
- Quadratic = x^2 + 3x + 2
- ( Factorise )
- ( x + 1 ) ( x + 2 )
- p( x ) = ( x - 2 ) ( x + 1 ) ( x + 2 )
3
Q
Fully factorise f( x ) = 2x^3 - 3x^2 + x - 6
A
- ( Firstly find a value which makes f( x ) = 0 )
- When x = 2, f( x ) = 0
- So ( x - 2 ) is a factor
- x - 2 | 2x^3 - 3x^2 + x - 6 | ( Dividing )
- ( 2x^2 + x + 3 ) ( x - 2 )
- ( Not possible to factorise )
- Hence f( x ) = ( 2x^2 + x + 3 ) ( x - 2 )