Polynomials


   
 
Factor Theorem
Statement
If p(x), a polynomial in x is divided by x-a and the remainder = f (a) is zero, then (x-a) is a factor of p(x).
 
Proof:
 
When p(x) is divided by x-a,
 
R = p(a) (by remainder theorem)
 
p(x) = (x-a).q(x)+p(a)
 
(Dividend = Divisor x quotient + Remainder Division Algorithm)
 
But p(a) = 0 is given.
 
Hence p(x) = (x-a).q(x)
 
 
Conversely if x-a is a factor of p(x) then p(a)=0.
 
p(x) = (x-a).q(x) + R
 
If (x-a) is a factor then the remainder should be zero (x-a divides p(x)
 
exactly)
 
R=0
 
By remainder theorem, R = p(a)
 
p(a)=0
 
Illustrative Examples
 
Example 1:
 
Determine whether x-2 is a factor of x2-7x+10.
 
Suggested answer:
 
p(x) = x2-7x+10 is divided by x-2.
 
R = p(2)
 
= 4-14+10=0, R=0
 
 
Example 2:
 
Determine whether x-3 is a factor of x3-3x2+4x-12.
 
Suggested answer:
 
p(x) = x3-3x2+4x-12 is divided by x-3.
 
R = p(3)
 
= 33- 3 x 32 + 4 x 3-12
 
= 27-27+12-12=0
 
 
Example 3:
 
Show that x+1 is a factor of 2x3+5x2-9x-12.
 
Suggested answer:
 
Let p(x) = 2x3+5x2-9x-12
 
when p(x) is by x+1 [x- (-1)],
 
R = p(-1)
 
= 2(-1)3+5(-1)2-9(-1)-12
 
= -2+5+9-12
 
= -14+14 = 0
 
R = 0
 
 
Example 4:
 
Find a so that x4+2x3-ax2+x-2 has (x+2) as its factor.
 
Suggested answer:
 
Let p(x) = x4+2x3-ax2+x-2
 
when p(x) is by x+2 [x- (-2)],
 
R = p(-2)
 
= (-2)4+2(-2)3-a(-2)2-2-2
 
= 16-16-4a-4
 
= -4a - 4
 
As (x+2) is a factor, R=0
 
-4a - 4 = 0
 
a = -1
 
Example 5:
 
If (x-2) and (x-3) are factors of x3+ax2+bx+12, find a and b.
 
Suggested answer:
 
Let p(x) = x3+ax2+bx+12
 
when p(x) is x-2,
 
 
R = p(2)
 
= 23+a(2)2+b(2)+12
 
= 8+4a+2b+12
 
= 20+4a+2b
 
But x-2 is a factor,
 
20+4a+2b = 0
 
 
 
when p(x) is by x-3,
 
R = p(3)
 
= 27+9a+3b+12
 
= 39+9a+3b
 
But x-3 is a factor,
 
 
39+9a+3b=0
 
[9a+3b=-39] 3
 
3a+b = -13 …(2)
 
Subtracting equation (1) from (2), we get,
 
3a+b-2a-b = -13+10
 
 
a = -3
 
Substitute a = -3 in equation (1),
 
2a+b = -10
 
-6+b = -10
 
b = -4
 
Example 6:
 
Factorise using factor theorem.
 
a) 2x2+x-3
 
b) 5x2+6x-8
 
Suggested answer:
 
a) 2x2+x-3
 
Let p(x) = 2x2+x-3
 
The factors of the constant 3 are +1, -1, +3, -3.
 
Let us find p(1) = 2+1-3 = 0
 
 
Divide 2x2+x-3 by x-1 to get the other factor.
 
 
Hence 2x2+x-3 = (2x+3) (x-1) .
 
b) 5x2+6x-8
 
Suggested answer:
 
Let p(x) = 5x2+6x-8
 
Factors of the constant '8' are 1,-1,2,-2,4, -4 and 8,-8.
 
Let us find the value of p(x) at each one of them to get one factor.
 
 
p(-2) = 20-12-8 = 0
 
 
Divide 5x2+6x-8 by x+2 to get the other factor.
 
 
 
 
     
   
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