Question

Factories 15.(2a+3b)²-4c²​

Answer 1

Step-by-step explanation:

(2a+3b)²-4c²

(4a²+9b²+12ab-4c²)=0

4a²+9b²+12ab-4c² =0 is the answer of this question

Answer 2

Solution :-

$\sf \red {:\implies (2a+3b)²-4c²}\\\\\sf :\implies {(2a)² + 2 \times 2a \times 3b +(3b)² }-4c²\\\\\sf \green{∵(x+y)²=x²+y²+2xy}\\\\ \sf :\implies 4a² +12ab +9b²-4c²\\\\\sf \red{:\implies 4a² +9b²-4c²+12ab}$

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Method of Common Factors:-

• In this method, we have to write the irreducible factors of all the terms
• Then find the common factors amongst all the irreducible factors.
• The required factor form is the product of the common term we had chosen and the left over terms.

Factorisation by Regrouping Terms:-

• Sometimes it happens that there is no common term in the expressions then
• We have to make the groups of the terms.
• Then choose the common factor among these groups.
• Find the common binomial factor and it will give the required factors.

Factorization by using Identities:-

• Sometimes the above methods are not helpful in Factorization. In such cases we use method of factorization using Identities.

$\sf{{\pmb{{\underline{Identities}}:}}}$

• $\sf{(a+b)(a+b)=\green{{\underline{\underline{\pmb{a^{2}+2ab+b^{2}}}}}}}$

• $\sf{(a-b)(a-b)=\green{{\underline{\underline{\pmb{a^{2}-2ab+b^{2}}}}}}}$

• $\sf{(a+b)(a-b)=\green{{\underline{\underline{\pmb{a^{2}-b^{2}}}}}}}$

• $\sf{(x+a)(x+b)=\green{{\underline{\underline{\pmb{x^{2}+(a+b)x+ab}}}}}}$