Class 10: Remainder and Factor Theorems – Lecture Notes – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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 $5$ .

Answer:

$x+3=0 \Rightarrow x=-3$

It is given that the remainder is $5$ .

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

Hence

$b(-3)^3+9(-3)^2+4(-3)-10=5$

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

Example 2:

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

Answer:

$x-2=0 \Rightarrow x=2$

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.

Answer:

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

$\Rightarrow (1)^3+a(1)^2+b(1)-45=0$

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

Similarly,

$(x+5)$ 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:

$x^3+13x^2+31x-45$

To factorize:

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

$x^2+14x+45$

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