The Factor Theorem
Introduction
Hopefully an easy ride. Started off with a quadratic equation x² - 8x - 20 = 0
This works out to
(x + 10)(x -2) = 0
So (x + 10) and (x -2) are factors and we can see either x = -10 and x = 2 are known as solutions. The take away was given we know solution we can reverse engineer factors. So if x = 2 is a solution, (x -2) is a factor
Detail
So here is comes, a long statement
If f(x) is a polynomial and f(a) = 0, then (x-a) is a factor of f(x) - argghhhhhh
The robot translated this into Iain English
If you plug a number into a polynomial and the result is zero, then (x − that number) must be one of its factors.
So an example question might be show that (x+5) is a factor of 2x³ + 15x² + 27x + 10.
To is x + 5 is a factor, it means a solution is x = -5 so
f(-5) = 2(-5)³ + 15(-5)² + 27(-5) + 10. f(-5) = -250 + 175 -135 + 10 f(-5) = 0
Another example
The video suggested when choose a possible solution to look at the factors of the constant so given
x³ + 4x² - 11x - 30
So look at the factors of 30, 1,2,3,5,6,10,15,30 and negatives. Using 3 for example
f(3) = 3³ + 4*3² - 11*3 - 30
f(3) = 27 + 36 - 33 - 30
= 0
(x-3) is a factor
Back to long Division
Once you know one factor, in the is (x-3) we can do long division to find the other.
x² + 7x + 10
----------------------
(x-3) | x³ + 4x² - 11x - 30
- (x³ - 3x²)
--------------------
7x² - 11x - 30
7x² - 21x
-----------
10x - 30
10x - 30
--------
0
So we now have
x³ + 4x² - 11x - 30 = (x-3)(x² + 7x + 10)
Where (x-3) is a linear and x² + 7x + 10 is a is a quadratic which we can factorise to
= (x-3)(x +5)(x +2)
Another Question
Show that
(2x+1) is a factor of 2x³ + 11x² + 17x + 6. Solving 2x + 1 = 0 2x= -1 x = -1/2
So
f(-1/2) = 2(-1/2)³ + 11(-1/2)² + 17(-1/2) + 6.
= 2(-1/8) + 11(1/4) + 17(-1/2) + 6
= -1/4 + 11/4 - 34/4 + 24/4
= 0