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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  $5$.

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 $a$.

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

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,

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

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

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

$x^2+14x+45$

$x-1 ) \overline {x^3+13x^2+31x-45}$

$x^3-x^2$

$\underline {- \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

$14x^2+31x-45$

$14x^2-14x$

$\underline {- \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

$45x-45$

$45x-45$

$\underline {- \ \ \ \ + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$

$\times$

Therefore

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

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

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

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