Factorisation of Algebraic expressions

Factorisation of Quadratic Polynomials of the Form ax2 + bx + c Using factor Theorem

Let us start with a simple question.

What is the value of (203)^{3}?

Yes, its value is 8365427.

For sure, you would have used the identity (*a* + *b*)^{3} = *a*^{3} + *b*^{3} + 3*ab* (*a* + *b*)

We can write (203)^{3} as (200 + 3)^{3} and then using the identity for (*a* + *b*)^{3}, we can find its value as this is an easier method as compared to multiplication.

However, we have many more applications of this identity. We can also factorise algebraic expressions using this identity.

For this, we have to rewrite the identity as follows:

This form of the identity is used to factorise the expressions of the form *a*^{3} **+** *b*^{3}.

In the same way, we can write the identity for *a*^{3} − *b*^{3} as follows:

To understand how to use these identities to factorise expressions, let us see an example.

Let us factorise the expression *x*^{6} − 729*y*^{6}.

*x*^{6} − 729*y*^{6}

= (*x*^{3})^{2} − (27*y*^{3})^{2}

= (*x*^{3} + 27*y*^{3}) (*x*^{3} − 27*y*^{3})[Using *a*^{2} − *b*^{2} = (*a* + *b*) (*a* − *b*)]

= [(*x*)^{3} + (3*y*)^{3}] [(*x*)^{3} − (3*y*)^{3}]

Using identities (1) and (2), we obtain

⇒ (*x* + 3*y*) (*x*^{2} + 9*y*^{2} − 3*xy*) (*x* − 3*y*)(*x*^{2} + 9*y*^{2} + 3*xy*)

This is the factorised form of the given expression.

To understand this method more clearly, let us solve some more examples.

**Example 1:**

Factorise the expression: 125*a*^{6} − 343

**Solution:**

125*a*^{6} − 343 = (5*a*^{2})^{3} − (7)^{3}

= (5*a*^{2 }− 7) (25*a*^{4} + 49 + 35*a*^{2}) [Using *a*^{3} − *b*^{3} = (*a* − *b*) (*a*^{2} + *b*^{2} + *ab*)]

This is the factorised form of the given expression.

**Example 2:**

Factorise the expression: *x*^{3} + 64*y*^{3} + 12*x*^{2}*y* + 48*xy*^{2} − (4*x* + *y*)^{3}

**Solution:**

*x*^{3} + 64*y*^{3} + 12*x*^{2}*y* + 48*xy*^{2} − (4*x* + *y*)^{3}

= *x*^{3} + (4*y*)^{3} + 3 (*x*) (4*y*) (*x* + 4*y*) − (4*x* + *y*)^{3}

= (*x* + 4*y*)^{3} − (4*x* + *y*)^{3} [Using (*a* + *b*)^{3} = *a*^{3} + *b*^{3} + 3*ab* (*a* + *b*)]

= (*x* + 4*y* − 4*x* − *y*) [(*x* + 4*y*)^{2} + (4*x* + *y*)^{2} + (*x* + 4*y*) (4*x* + *y*)]

[Using *a*^{3} − *b*^{3} = (*a* −* b*) (*a*^{2} + *b*^{2} + *ab*)]

= (− 3*x* + 3*y*) [*x*^{2} + 16*y*^{2} + 8*xy* + 16*x*^{2} + *y*^{2} + 8*xy* + (4*x*^{2} + 17*xy* + 4*y*^{2})

= 3 (*y* − *x*) [21*x*^{2} + 21*y*^{2} + 33*xy*]

= 3 (*y* − *x*) × 3 (7*x*^{2} + 7*y*^{2 }+ 11*xy*)

= 9 (*y* − *x*) (7*x*^{2} + 7*y*^{2} + 11*xy*)

This is the factorised form of the given expression.

**Example 3: **

Factorise the following expressions.

(1)

(2)

(3)

(4)

**Solution:**

**Factorisation of Quadratic Polynomials**

We know that 7 × 6 = 42. Here, 7 and 6 a…

To view the complete topic, please