Polynomials of degree 3 or higher are difficult to factorize, so long division or synthetic division techniques are used to solve this problem. These methods are quite popular and frequently used by students during their early school studies. However, students are always in the need of a shortcut. This is where the concept of remainder theorem came into existence.

Once you will complete this article, you will get to know how you can use the remainder theorem to factorize the polynomial and what are the applications in the real-life. This is true that remainder theorem does not give you much information like other theorems, still an effective style to continue with the factorization.

Before we continue ahead, let us consider a simple example of listing factors of 27. If you would look closely, probably the factors are 1,3,9, and 27 and the remainder is zero. The method is simpler in case of an integer but what is the solution if you are using the polynomials. Other than these, if you are going to divide 27 with any other number then it will leave a remainder. For example, if you will divide the 27 with 6 then it will give a remainder of 3.

Finding factors for a quadratic polynomial is little more difficult but quickly solved by Algebra students with the right practice and technique. What about the equations with degree 3 or higher? Simply, the lead term higher than 2 as shown in the example below –

x^{3} + 2x^{2} – 11x – 12

In this case, the better idea is to use the long division or synthetic method to factorize the Polynomials that are highly effective and alternatives techniques are always available.

**Remainder Theorem**

Let us first discuss the definition of the Remainder Theorem that states that if we are dividing a polynomial function f(x) by (x – h), then the remainder is f(h). Here, when you know that the remainder is f(h) then we don’t have to use any other techniques, just check when x comes equal to h to calculate the remainder. When h is the real number, we will get a quotient too with the name p(x) and the theorem could be represented as –

f(x)/(x – h) = p(x) with a remainder of r

Like we did with our simple division problems in integers, we can manipulate this function to get the following.

f(x) = (x – h) * p(x) + r

you don’t have to go into much deeper as what is derivation or proof the poem. You just need to know how you can use it for the real-world applications. The remainder is always smaller than the number you divide by. If we are dividing by the linear factor then the remainder would always be a constant value. In most of the cases, the remainder is the number when divided by the term “x-a” in the mathematics.

**Example:**** Find the root of the polynomial \(x^{2}- 3x – 4\).**

**Solution:** \( x^{2} – 3x – 4\)

\( f(4) = 4^{2} – 3(4) – 4\)

\( f(4) = 16 – 16 = 0\)

So, (x-4) must be a factor of \( x^{2} – 3x – 4\)

**Example:** **Find the remainder when\(t^{3} – 2t^{2} + t + 1\) is divided by t – 1.**

**Solution:** Here, \(p(t) = t^{3} – 2t^{2} + t + 1\)

∴ p (1) = (1)3 – 2(1)2 + 1 + 1= 2

By the Remainder Theorem, 2 is the remainder when \(t^{3} – 2t^{2} + t + 1\)