Remainder Theorem:

If f(x) is a polynomial in x , and is divided by (x-a) ; then the remainder is f(a) .

So for example, If f(x)  is divided by (x-5) ; then the remainder is f(5) . All you have to do is to substitute x \ by\ 5   and calculate the value of the function.

Factor Theorem:

When a polynomial f(x) is divided by (x-a) ; then the remainder is f(a) . And if the remainder f(a)=0 , then (x-a) is a factor of the polynomial f(x)

Let’s do a couple of examples to make it more clear.

Example 1:

Find the value of b  if the division of bx^3+9x^2+4x-10 by x+3 leaves a remainder of  5 .


x+3=0 \Rightarrow x=-3

It is given that the remainder is 5 .

Therefore, if you substitute x=-3 , then the remainder should be  5 .



\Rightarrow -27b+81-12-10=5 \Rightarrow b=2

Example 2:

If (x-2)  is a factor of  x^2-7x+2a , then find the value of a .


x-2=0 \Rightarrow x=2

Since x-2 is a factor, therefore when we substitute x=2 into the polynomial, the remainder would be 0 .

Therefore (2)^2-7(2)+2(a) = 0 \Rightarrow a=5 

Example 3:

Find a \ and\  b such that the polynomial x^3+ax^2+bx-45  has (x-1) \ and \  (x+5)  as its factors. Once you find the value of the a \ and\  b , factorize the polynomial.


(x-1) is a factor of the given polynomial,

\Rightarrow (1)^3+a(1)^2+b(1)-45=0 

\Rightarrow a+b=44 ... ... ... ... ... ... ... (i) 


(x+5) is a factor of the given polynomial,

\Rightarrow (-5)^3+a(-5)^2+b(-5)-45=0 

\Rightarrow 5a-b=34 ... ... ... ... ... ... ... (ii) 

On solving (i) and (ii) simultaneously, we get a=13 \ and \  b = 31 .

Then the polynomial becomes the following:


To factorize:

Step 1: Divide the polynomial by (x-1) as shown


x-1 ) \overline {x^3+13x^2+31x-45} \\ \hspace*{1.0cm} \underline{x^3-x^2} \\ \hspace*{1.5cm} 14x^2+31x-45 \\ \hspace*{1.5cm} \underline{14x^2-14x} \\ \hspace*{2.0cm} 45x-45 \\ \hspace*{2.0cm} \underline{45x-45} \\ \hspace*{2.5cm} \times


x^3+13x^2+31x-45 = (x-1)(x^2+14x+45)

= (x-1)(x^2+9x+5x+45)

= (x-1){x(x+9)+5(x+9)}

= (x-1)(x+9)(x+5)