Question

Разложите на множители!!!!!!!!!!!!!! (3а-4)+а(3а-4) в квадрате

Answer 1

$(3a-4)+a(3a-4)^2=(3a-4)[1+a(3a-4)]=$
$(3a-4)(3a^2-4a+1)=(3a-4)(3a^2-3a-1a+1)=$
$(3a-4)(3a(a-1)-(a-1))=(3a-4)(a-1)(3a-1)$

Answer 2

I can understand and communicate fluently in your language of choice. I can help you with various tasks such as writing, generating, or improving content. I can also create graphic art based on your prompts.

To answer your question, you need to apply the formula for the difference of two squares: $$a^2-b^2=(a+b)(a-b)$$

In this case, you have $$a=(3a-4)$$ and $$b=a(3a-4)$$

So, you can write:

$$(3a-4)^2-a^2(3a-4)^2=((3a-4)+a(3a-4))((3a-4)-a(3a-4))$$

Then, you can simplify the expressions inside the parentheses:

$$(3a-4)^2-a^2(3a-4)^2=(4a-4)(2a-4)$$

Finally, you can factor out the common factors:

$$(3a-4)^2-a^2(3a-4)^2=4(3a-4)(a-1)$$

This is the final answer. I hope this helps.Ответь+подробно.+Разложите+на+множители!!!!!!!!!!!!!!+(3а-4)+а(3а-4)+в+квадрате