In Class 8 we studied basic concepts. You can do a quick revision of basic concepts before you start. As you read the following section, you will realize that the formulas that you learnt in Expansions will be used extensively. It might be a good idea to quickly brush up the formulas as well.

There are various ways of factoring a given polynomial. We will list those down with one example each.

(i) Factorization of algebraic expression by taking out common factors: Find the HCF of all the terms of the polynomial and take that out as a common factor.

Example: $18x^3y - 45x^2yz = 9x^2y(2x-5z)$. HCF of the two terms $18x^3y$ and $45x^2yz$ is $9x^2y$

(ii) Factorization by grouping of terms: Arrange the terms in such a way that there is a common factor for each group. Continue this process until the polynomial is completely factorized.

Example: $xy-ab+bx-ay = xy+bx - ab - ay = x(y+b)-a(y+b) = (y+b) (x-a)$

(iii) Factorization by making a perfect square: At times, trinomial can be written as a perfect square of a binomial expression. The following formulas can be effectively used.

${(a+b)}^2=\ a^2+2ab+b^2$

${(a-b)}^2=\ a^2-2ab+b^2$

${(a+b+c)}^2=\left(a^2+b^2+c^2\right)+2(ab+bc+ca)$

Example: $1-8x+16x^2 = 1 - 2 \times 8x + (4x)^2 = (1-4x)^2$

(iv) Factorizing the difference of two squares:  Sometimes, the expressions can be written as a difference of the squares. In that case you can use the following formula.

$a^2-b^2 = \left(a+b\right)\left(a-b\right)$

Example: $9x^2-4y^2 = (3x)^2 - (2y)^2 = (3x + 2y)(3x-2y)$

(v) Factorizing the sum and difference of two cubes: Sometimes, the expressions can be written as a sum or difference of the squares. In that case you can use the following formulas.

$a^3+b^3 = (a+b)(a^2 -ab + b^2)$

$a^3-b^3 = (a-b)(a^2 +ab + b^2)$

Examples:

$a^3 + 27 = (a)^3 + (3)^3 = (a+3)(9-3a+9)$

$a^3 - 27 = (a)^3 - (3)^3 = (a-3)(9+3a+9)$

(vi) Factorization by using the formula for the cube of a binomial: In this case, we can use the following formula.

${(a+b)}^3=\ a^3+b^3+3ab(a+b)$

${(a-b)}^3=\ a^3-b^3-3ab(a-b)$

Examples:

$8a^3+b^3+12a^2b+6ab^2 = (2a)^3 + (b)^3 +6ab(2a+b) = (2a)^3 + (b)^3 +3 \times 2a \times b(2a+b) = (2a+b)^3$

$8a^3-b^3-12a^2b+6ab^2 = (2a)^3 - (b)^3 -6ab(2a-b) = (2a)^3 - (b)^3 -3 \times 2a \times b(2a-b) = (2a-b)^3$

(vii) Factorization of a quadratic polynomial by splitting the middle term: For a quadratic expression $x^2+bx+c =0 \Rightarrow x^2 + (p+q)x + pq=0 \Rightarrow (x+p)(x+q) = 0$ where $p+q = b$ and $pq = c$.

Example:

$x^2 + 6x + 8 = \Rightarrow x^2 + (4+2)x + 4 \times 2 = \Rightarrow (x+4)(x+2)$

$x^2-7x+12 = \Rightarrow x^2 +(-3-4)x + (-3) \times (-4) = \Rightarrow (x-3)(x-4)=$

(viii) Factorization of expressions which are factorizable by splitting the middle term: Sometimes the polynomial is not a quadratic polynomial. We can factorize the polynomial by splitting the middle term such that their sum is equal to the middle term and the product is equal to the product of first and last term. This sounds a little complex but the following example will make it easier.

Example: $x^2y^2-xy-72 = x^2y^2 -9xy+8xy-72=xy(xy-9)-8(xy-9) = (xy-9)(xy-8)$