Question

$a {}^{4} + a {}^{2}b {}^{2} - 72b {}^{4}$
plz solve this step by step plz ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: {a}^{4} + {a}^{2} {b}^{2} - 72 {b}^{4}$

To factorize this, we have to use the concept of Splitting of middle terms.

So, above expression can be rewritten as

$\rm \: = \: {a}^{4} + 9{a}^{2} {b}^{2} - 8 {a}^{2} {b}^{2} - 72 {b}^{4}$

$\rm \: = \: {a}^{2}( {a}^{2} + 9 {b}^{2}) - 8 {b}^{2}( {a}^{2} + 9 {b}^{2})$

$\rm \: = \: ( {a}^{2} + 9 {b}^{2})( {a}^{2} - 8 {b}^{2})$

Hence,

$\boxed{\rm{ \: {a}^{4} + {a}^{2} {b}^{2} - 72 {b}^{4} \: = \: ( {a}^{2} + 9 {b}^{2})( {a}^{2} - 8 {b}^{2}) \: }}$

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Basic Concept

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to find numbers m and n such that m + n = b and mn = ac.

After finding m and n, we split the middle term in the given quadratic expression as mx + nx and get required factors by grouping the terms.

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Some useful Identities

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

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

$\boxed{\sf{ \: {a}^{2} - {b}^{2} = (a + b)(a - b) \: }}$

$\boxed{\sf{ \: \: (x + a)(x + b) \: = \: {x}^{2} + (a + b)x + ab \: \: }}$

Answer 2

Answer :

Therefore ,

$\sf \: ({a}^{2} + {9b}^{2})( {a}^{2} - {8b}^{2} )$

Step-by-step-Explaination :

Here, the given expression is :-

Step 1 :

First Spilit into Middle terms factors .

$\sf \: {a}^{4 } + {a}^{2} {b}^{2} - 72 {b}^{4}$

Now , to factorize it , We need to know a concept known as Splitting of Middle terms . Here , we take a factor by which dividing of 72 given the two same factors needed for a^2 b^2 by their substraction .

Hence,

Factors of 72 :-

$\sf \: 1 \times 72 = 72$

$\sf \: 2 \times 36 = 72$

$\sf \: 3 \times 24 = 72$

$\sf \: 4\times 18= 72$

$\sf \: 6\times 12= 72$

$\sf \: 9\times 8= 72$

Here , all the factors are given :-

• 1
• 2
• 3
• 4
• 6
• 8
• 9
• 12
• 18
• 24
• 36
• 72

But , see here carefully 9 and 8 make a difference of 1 . So, we will take it :-

$\longrightarrow \sf \: {a}^{4 } + {9a}^{2} {b}^{2} - 8 {a}^{2} {b}^{2} - 72 {b}^{4}$

Step 2 :

Now taking common ,

$\longrightarrow \sf \: {a}^{2} ( {a}^{2} + {9b}^{2}) - 8 {b}^{2} ( {a}^{2} + 9{b}^{2} )$

Now , see always you need to check that here , inside the brackets are same or not if same , it means you are right,

$\longrightarrow \sf \: ( {a}^{2} + {9b}^{2}) ( {a}^{2} - {8b}^{2} )$

Therefore,

$\longrightarrow \sf \: {a}^{4 } + {a}^{2} {b}^{2} - 72 {b}^{4} = ( {a}^{2} + {9b}^{2}) ( {a}^{2} - {8b}^{2} )$

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☆Additional Information :

☆ This is done from the chapter Factorisation.

☆ The process of finding two or more expressions whose products is the given expression is called Factorisation.

☆ Method of Factorisation:-

1. Taking out common factors .
2. Grouping .
3. Using algebraic identities

The identities are as follows given in the attachment kindly check it .

☆ HCF of two or more polynomial is the largest common factor of the given polynomials .

Factorisation of Trinomials :

Case I :

When the trinomial is of the form x^2 + px + q , where p and q are integers .

• Let x^2 + px + q = (x+a)(x+b) = x^2 +(a+b)x + ab .

Thus , if we want to factorise the trinomial of the form x^2 + px + q , we need to find two integers a and b such that a + b = q and ab = q.

Therefore ,

Split p , ( the coefficient of x ) into two parts such that the algebraic sum of these two parts is p and their products q .

Case II :

When the trinomial is of the form ax^2 + bx + c , where a , b and c are integers .

We want to find two integers A and B such that

A + B = b and AB = ac

Therefore ,

Split b ( the coefficient of x ) into two parts such that the algebraic sum of these two parts is b and their products is ac .