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Factorisation

Factorisation Using Division

Factorisation is the process of reducing the bracket of a quadractic equation, instead of expanding the bracket and converting the equation to a product of factors which cannot be reduced further. For example, factorising (x²+5x+6) to (x+2) (x+3). Here, (x+2) (x+3) is factorisation of a polynomial (x²+5x+6). These factors can be either variable, integers or algebraic expressions. Basically, Factorisation is the reverse function of multiplication. A form of disintegration, factorisation entails the gradual breakdown of a polynomial into its factors. Let us see how to factorise polynomials and what are different types of factorisation.

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What is Factorisation?

Factorisation is disintegrating a composite number into its factors. Basically a form of division, factorisation entails finding numbers which give remainder as zero. In the case of composite numbers the calculation seems simple, but when it comes to factorisation of polynomials we often get confused. The process needs immense understanding and practice.

Factorisation

While factorizing polynomials using division method we must keep the following points in mind:

  • Finding factors of a polynomial expression by division method is just like doing any simple division, the only thing to be kept in mind is the accuracy of variables and coefficients.
  • Factorisation by division method is the conventional method of finding factors of a polynomial expression.

Factorisation of polynomials can be done in two ways. One by normal division and second by long division method.

Browse more Topics under Factorization

Factorisation by Division

In factorisation by simple division method, we first break the polynomial into its direct factors. For example, if we divide 8y3+7y2+6y by 2y, we first break the polynomial into its basic factors, i.e : 2y(4y) 2+ 2y (7/2 *y) + 2y(3)

Next, we write the common factor separately, where we get: 2y { 4y2+(7/2y) + 3}. In the last step, we divide the expression as asked in the question i.e: 2y {4y2+(7/2y) + 3} / 2y. The answer to this shall be: 4y2+ (7/2y) + 3

Example 1: Divide 16(x2yz + xy2z+xyz2) by 4xyz

Solution : 2×2×2×2× [(x×x×y×z) + (x×y×y×z) + (x×y×z×z)]
= 2×2×2×2× {x×y×z (x+y+z)}
= 16xyz (x+y+z)
Now divide the polynomial as given in the question:
= 4*4xyz (x+y+z) / 4xyz
= 4(x+y+z)

Finding Factors: The Long Division Way

While finding factors of a polynomial using division method we need to accurately follow the steps given underneath:

  • Firstly, we arrange the polynomials in descending order. Wherever a term is missing we replace it with a zero ( (0). For example: Take a polynomial : x3+6x2+12+3x / x+3. Here, we first rearrange the polynomial as x3+6x2+3x+12
  • When we start the division of a polynomial, our first target is the first term of the polynomial. We divide the dividend’s first term with the first term of the divisor. From here we get our first quotient:

x+3 |x3+6x2+12| x2                       ( x3 ÷ x = x2)

  • Then we multiply this quotient with the divisor, in the example we get:

(x+3) × x2 = x3+3x2

  • We now subtract the product from the dividend, like we do in the normal division calculations. Whatever the difference we get shall be our next dividend.

x+3 |x3+6x2+12| x2    gives    x+ 3x2 

  • Bring down the next term and whatever the answer we get here is again divided by the divisor in a similar manner. We repeat the steps until we get a remainder which is lower than the divisor or is a Zero.

Now, if the remainder is a zero, we come to the conclusion that the divisor is the factor of the given polynomial.

Learn more about Factorisation using identities.

 

Solved Examples For You

Question 1: Let f(x)=2x3+16x2+44x+42 be a polynomial having one of the factors as (x2+5x+7), then the other factor of f(x) would be a multiple of:

A) 1      B) 2      C) 3      D) 4

Answer : B) Since f(x) is a cubic polynomial, and one of the factors is a polynomial of degree two, then we can say that the other factor will be a polynomial of the form ax + b;  where ‘a’ nd ‘b’ are two constants and a ≠ 0. Hence, we can write:

2x3+16x2+44x+42 = (x2+5x+7) × (ax + b) = ax3 + bx2 + 5ax2 + 5bx + 7ax + 7b
or 2x3+16x2+44x+42 = ax3 + (b + 5a)x2 +x2 + (5b + 7a)x + 7b

Compairing the coefficients of x on both sides, we have 2 = a and 42 = 7b. Therefore, b = 6 and a = 2. hence the other factor is 2x + 6 or 2(x+3) which is a multiple of 2.

Question 2: What are the various types of factoring?

Answer: The six types of factoring are: greatest common factor, difference in two squares, grouping, sum or difference in two cubes, trinomials, and general trinomials.

Question 3: What is meant by factorisation?

Answer: Factorisation refers to the breaking up of a number into smaller numbers that on multiplication will provide you with the original number. Factorisation means the splitting up of a number into various factors or divisors.

Question 4: What is the use of factorisation in real life?

Answer: Factorisation has many uses in real life, some of which are:  division of something into equal pieces, exchange of money, comparing different prices, doing time management, and making calculations while travelling.

Question 5: Explain the three steps for solving quadratic equation by factoring?

Answer: The three steps for solving quadratic equations by factoring are as follows:

  • Step 1: Writing equation in the correct form. One would need to set the equation equal to zero such that terms are in descending order.
  • Step 2: Use various factoring strategies to factor the problem.

Step 3: Make use of the Zero Product Property and set each factor that has a variable equal to zero.

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