Factorization by Splitting the Middle Term: The method of Splitting the Middle Term by factorization is where you divide the middle term into two factors. We know that composite numbers can be expressed as the product of prime numbers. For example, \(42 = 2 \times 3 \times 7\). Here, \(2,\,3\) and \(7\) are the prime factors of \(42\).

Similarly, algebraic expressions can also be expressed as the product of irreducible factors. Thus, the process of finding two or more expressions whose product is the given expression is called factorisation. There are many ways to factorise a polynomial, like taking out the common factors, grouping the terms, splitting the middle term, etc. In this article, we will have a detailed discussion about the method of factorisation by splitting the middle term.

Factorisation of Polynomials by Splitting the Middle Term

We know that the process of writing an algebraic expression as a product of its factors is called factorisation. A polynomial can be factorised using various methods. Below are the mentioned ways to factorise a polynomial.

1. Factorisation using Common Factors
2. Factorisation by Regrouping
3. Factorisation using Identities
4. Factorisation by splitting the Middle Term
5. Sum or Difference of Two Cubes

We will learn about factorisation by splitting the middle term in detail.

Factorisation by Splitting the Middle Term

In this section, we shall learn the factorisation of trinomials of the form \(a{x^2} + bx + c\) where \(a,\,b\) and \(c\) are real numbers.

Rule to factorise trinomial \(a{x^2} + bx + c\) where \(a,\,b\) and \(c\) are real numbers, is to split \(b\) (the coefficient of \(x\)) into two real numbers such that the algebraic sum of these two numbers is \(b\) and their product is \(c\), then factorise by grouping method.

It is not always possible to factorise a trinomial \(a{x^2} + bx + c\) (i.e., a quadratic expression); the following rule can save a lot of time:

For the expression \(a{x^2} + bx + c\), work out \({b^2} – 4ac\). If it is a perfect square, then the given expression will factorise; otherwise, not.

Let us look at a couple of examples.

Example 1: Factorise the following trinomial.

\(2{x^2} – 7x – 15\)

Solution: To factorise \(2{x^2} – 7x – 15\), we want to find two real numbers whose sum is \(b = – 7\) and the product is \(ac = 18\). By trial, we see that \(( – 10) + 3 = – 7\) and \(( – 10) \times 3 = – 30\).

Therefore, \(2{x^2} – 7x – 15 = 2{x^2} – 10x + 3x – 15\)

\( = 2x\left( {x – 5} \right) + 3\left( {x – 5} \right)\)

\( = \left( {x – 5} \right)\left( {2x + 3} \right)\)

Example 2: Factorise \({x^2} – 8x + 12\)

The middle term is \(-8x\). We have to split \(-8x\) so that the obtained numbers, when multiplied together, give \(12\).

Thus \(-8x\) can be written as \( – 2x – 6x\)

Therefore, \({x^2} – 8x + 12 = {x^2} – 2x – 6x + 12\)

\( = x(x – 2) – 6(x – 2)\)

\( = (x – 6)(x – 2)\)

Let us learn the concept in an elaborated way.

We know that the expansion of the product of the linear expressions is a quadratic expression.

For example,

\((2x + 7)(3x – 5) = 2x(3x – 5) + 7(3x – 5) \ldots ({\rm{i}})\)

\( = 6{x^2} – 10x + 21x – 35 \ldots (ii)\)

\( = 6{x^2} + 11x – 35 \ldots (iii)\)

The factorisation is the reverse process of expansion. By reversing the process above, we factorise the quadratic expression into a product of two linear factors. Thus, we say the factorised form of \(6{x^2} + 11x – 35\) is \((2x + 7)(3x – 5)\).

Notice the above example again carefully with particular attention to equations \(\left( {{\rm{ii}}} \right)\) and \(\left( {{\rm{iii}}} \right)\). To factorise the expression \(6{x^2} + 11x – 35\), a trinomial, we split the middle term \(11x\) into two terms: \( – 10x + 21x\).

This process can be called the factorisation of a trinomial by splitting the middle term to group the common terms.

Let us look at a few examples to become familiar with using discs to represent numbers and unknown quantities and thus understand the process with the help of a model.

The discs can be cut from stiff paper and used as algebra manipulatives, or different coloured circular bindis can be used instead.

For example, to factorise, \({x^2} + 7x + 10\), we arrange the discs representing \({x^2} + 7x + 10\) to form a rectangular array, as shown below.

Now, observe the pattern of the arrangement of the discs. The rectangle is formed first by arranging the \({x^2}\) discs at the top left corner, and the \(10\) discs of \(1\) at the bottom right corner as shown in the first figure. Then we put in the \(x\) discs to complete the rectangle, as shown in the second figure.

The two linear factors are, therefore, \((x + 2)\) and \((x + 5)\), as shown in figure \(3\).

Observe that the constant term \(10\) is factorised into \(2 \times 5\) and the \(7x\)-discs, i.e., \(7x\) can be divided into two groups (\(2x\) and \(5x\)) to complete the rectangle. Thus, we have,

\({x^2} + 7x + 10 = {x^2} + 5x + 2x + 10\)

\( = x(x + 2) + 5(x + 2)\)

\( = (x + 2)(x + 5)\)

Let us look at another way to factorise the trinomials with the help of a multiplication frame to help us factorise the above expression. Let us learn this method step by step.

Step 1: First we write \({x^2}\) in the top-left corner and \(10\) at the bottom-right corner of the frame.

Step 2: Consider the factors of \({x^2}\) and \(10\), and write them in the appropriate corner and row. Observe that for the factors of \({x^2}\), we can write \(x \times x\). For factors of \(10\), we can write \(1 \times 10\) or \(2 \times 5\).

Step 3: Multiply the factors and check if the result matches the quadratic expression.

Consider the factors of \(1\) and \(10\). Multiply \(x\) by \(10\) and \(1\) by \(x\).

Check: \(x(10) + 1(x)\)

\( = 10x + x\)

\( = 11x \ne 7x\)

Hence, \({x^2} + 7x + 10 \ne (x + 1)(x + 10)\).

Let us consider factors \(2\) and \(5\).

Check if the sum of terms containing \(x\) is \(7x\) 

Check: \(x(5) + 2(x)\)

\( = 5x + 2x\)

\( = 7x\)

Therefore, \({x^2} + 7x + 10 = (x + 2)(x + 5)\).

Let us take and understand this concept better with the help of another illustration.

Factorise \({x^2} – 8x + 12\).

Step1: First we write \({x^2}\) in the top-left corner and \(12\) at the bottom right corner of the frame.

Step 2: Consider the factors of \({x^2}\) and \(12\). For factors of \(12\), we can write \( – 1 \times – 12,\, – 2 \times – 6\) and \( – 3 \times – 4\) since the coefficient of \(x\) is negative.

Step 3: Let us consider the factors of \(-1\) and \(-12\) Write them in the appropriate row and column. Multiply the factors and check if the sum of terms containing \(x\) is \(-8x\).

Check: \(x( – 12) – ( – 1)(x)\)

\( = – 12x + ( – 1x)\)

\( = – 13x \ne – 8x\)

Therefore, \({x^2} – 8x + 12 \ne (x – 1)(x – 12)\)

Check: \(x( – 4) – ( – 3)(x)\)

\( = – 4x + ( – 3x)\)

\( = – 7x \ne – 8x\)

Therefore, \({x^2} – 8x + 12 \ne (x – 3)(x – 4)\)

Check: \(x( – 6) – ( – 2)(x)\)

\( = – 6x + ( – 2x)\)

\( = – 8x = – 8x\)

Therefore, \({x^2} – 8x + 12 = (x – 2)(x – 6)\)

Factorisation of Polynomials by Splitting the Middle Term Questions

Q.1, Factorise the following trinomial.
\({x^2} + 9x + 18\)
Ans: To factorise \({x^2} + 9x + 18\), we want to find two real numbers whose sum is \(9\) and the product is \(18\). By trial, we see that \(3 + 6 = 9\) and \(3 \times 6 = 18\).
Therefore, \({x^2} + 9x + 18 = {x^2} + 3x + 6x + 18\)
\( = x(x + 3) + 6(x + 3)\)
\( = (x + 3)(x + 6)\)

Q.2. Factorise the following trinomial.
\(2{x^2} + 7x + 3\)
Ans: Step 1: First we write \(2{x^2}\) in the top-left corner and \(3\) at the bottom-right corner of the frame.

Step 2: For the factors of \(2{x^2}\), we can write \(2{x^2} = 2x \times x\) 
The factors of \(3\) are \(1\) and \(3\)
Write them in the appropriate column and row.

Step 3: Multiply the factors and check if the sum of the terms containing \(x\) is \(7x\). 
Check \(2x(1) + 3(x) = 2x + 3x\)
\(=5x\)
\(5x \ne 7x\)

Therefore, \(2{x^2} + 7x + 3 \ne (2x + 3)(x + 1)\)
Consider switching factors \(1\) and \(3\) to a different column and row.
Check \(2x(3) + 1(x) = 6x + x\)
\( = 7x\)
\(7x = 7x\)

Therefore, \(2{x^2} + 7x + 3 = (2x + 1)(x + 3)\).

Q.3. Factorise the following trinomial.
\(84 – 2p – 2{p^2}\)
Ans: We note that \(84 – 2p – 2{p^2} = 2\left( {42 – p – {p^2}} \right)\).
To factorise \(\left( {42 – p – {p^2}} \right)\), we want to find two real numbers whose sum is equal to \(-1\) and product is \(42 \times ( – 1) = – 42\). By trial, we see that \(( – 7) + 6 = – 1\) and \(( – 7) \times 6 = – 42\).
\( = 2\left( {42 – 7p + 6p – {p^2}} \right)\)
\( = 2[7(6 – p) + p(6 – p)]\)
\( = 2(6 – p)(7 + p)\).

Q.4. Factorise the following trinomial.
\(7\sqrt 2 {x^2} – 10x – 4\sqrt 2 \)
Ans: To factorise \(7\sqrt 2 {x^2} – 10x – 4\sqrt 2 \), we want to find two real numbers whose sum is equal to \(-10\) and product is \(7\sqrt 2 \times ( – 4\sqrt 2 ) = – 56\). By trial, we see that \(( – 14) + 4 = – 10\) and \(( – 14) \times 4 = – 56\).
Therefore, \(7\sqrt 2 {x^2} – 10x – 4\sqrt 2 = 7\sqrt 2 {x^2} – 14x + 4x – 4\sqrt 2 \)
\( = 7\sqrt 2 x(x – \sqrt 2 ) + 4(x – \sqrt 2 )\)
\( = (x – \sqrt 2 )(7\sqrt 2 x + 4)\)

Q.5. Factorise the following trinomial.
\({x^2} + \frac{1}{4}x – \frac{1}{8}\)
Ans: Since, \(\frac{1}{2} – \frac{1}{4} = \frac{1}{4}\) and \(\frac{1}{2} \times \left( { – \frac{1}{4}} \right) = – \frac{1}{8}\)
Therefore, \({x^2} + \frac{1}{4}x – \frac{1}{8} = {x^2} + \frac{1}{2}x + \frac{1}{4}x – \frac{1}{8}\)
\( = x\left( {x + \frac{1}{2}} \right) – \frac{1}{4}\left( {x + \frac{1}{2}} \right)\)
\( = \left( {x + \frac{1}{2}} \right)\left( {x – \frac{1}{4}} \right)\).

Summary

In this article, we first learnt the definition of factorisation in detail with the help of an example. Then, we learnt one of the methods to factorise polynomial, i.e., the method of splitting the middle term. We learnt two ways to factorise the polynomials with the help of examples, and later we solved some examples based on the concept of splitting the middle term to strengthen our grip over the concept.

FAQs on Factorization by Splitting the Middle Term

The frequently asked questions regarding Factorization by Splitting the Middle Term are given below:

Q.1. How do you divide the middle term in factorisation?
Ans: Rule to factorise trinomial \(a{x^2} + bx + c\), where \(a,\,b\) and \(c\) are real numbers, is to split \(b\) (the coefficient of \(x\)) into two real numbers such that the algebraic sum of these two numbers is \(b\) and their product is \(ac\), then factorise by grouping method.

Q.2. Define factorisation of polynomials.
Ans: When an algebraic expression can be written as a product of two or more algebraic expressions, each expression is called a factor of the given expression. Thus, finding two or more expressions whose product is the given expression is called factorisation.

Q.3. What is the easiest way to find the splitting of the middle term?
Ans: For the expression \(a{x^2} + bx + c\), work out \({b^2} – 4ac\). If it is a perfect square, then the given expression will factorise; otherwise, not. Split \(b\) (the coefficient of \(x\)) into two real numbers such that the algebraic sum of these two numbers is \(b\) and their product is \(ac\), then factorise by grouping method.

Q.4. How do you break the middle term of a quadratic equation?
Ans: The rule to factorise the trinomial \(a{x^2} + bx + c\), where \(a,\,b\) and \(c\) are real numbers, is to split \(b\) (the coefficient of \(x\)) into two real numbers such that the algebraic sum of these two numbers is \(b\) and their product is\(ac\).

Q.5. How do you factorise polynomials?
Ans: We factorise the polynomials by collecting the like terms together and then finding the common term between them. There are various methods to do the factorisation of a polynomial like
1. Factorisation using common factors
2. Factorisation by regrouping
3. Factorisation using identities
4. Factorisation by splitting the middle term
5. Sum or difference of two cubes

Study Factorisation by Direct Method and Grouping Here

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