Math, asked by sahar2, 1 year ago

please answer question with full explanation
factorize

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Answered by Adarshm
0




36{x}^{2}-49{y}^{2}-6x-7y36x​2​​−49y​2​​−6x−7y



 1

In general, given a{x}^{2}+bx+cax​2​​+bx+c, the factored form is:

a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a})a(x−​2a​​−b+√​b​2​​−4ac​​​​​)(x−​2a​​−b−√​b​2​​−4ac​​​​​)


2

 

In this case, a=36a=36, b=-6b=−6 and c=-49{y}^{2}-7yc=−49y​2​​−7y

36(x-\frac{6+\sqrt{{(-6)}^{2}-4\times 36(-49{y}^{2}-7y)}}{2\times 36})(x-\frac{6-\sqrt{{(-6)}^{2}-4\times 36(-49{y}^{2}-7y)}}{2\times 36})36(x−​2×36​​6+√​(−6)​2​​−4×36(−49y​2​​−7y)​​​​​)(x−​2×36​​6−√​(−6)​2​​−4×36(−49y​2​​−7y)​​​​​)


3

 

Simplify

36(x-\frac{6+6\sqrt{1+28y(7y+1)}}{72})(x-\frac{6-6\sqrt{1+28y(7y+1)}}{72})36(x−​72​​6+6√​1+28y(7y+1)​​​​​)(x−​72​​6−6√​1+28y(7y+1)​​​​​)


4

 

Factor out the common term 66

36(x-\frac{6(1+\sqrt{1+28y(7y+1)})}{72})(x-\frac{6-6\sqrt{1+28y(7y+1)}}{72})36(x−​72​​6(1+√​1+28y(7y+1)​​​)​​)(x−​72​​6−6√​1+28y(7y+1)​​​​​)


5

 

Simplify \frac{6(1+\sqrt{1+28y(7y+1)})}{72}​72​​6(1+√​1+28y(7y+1)​​​)​​ to \frac{1+\sqrt{1+28y(7y+1)}}{12}​12​​1+√​1+28y(7y+1)​​​​​

36(x-\frac{1+\sqrt{1+28y(7y+1)}}{12})(x-\frac{6-6\sqrt{1+28y(7y+1)}}{72})36(x−​12​​1+√​1+28y(7y+1)​​​​​)(x−​72​​6−6√​1+28y(7y+1)​​​​​)


6

 

Factor out the common term 66

36(x-\frac{1+\sqrt{1+28y(7y+1)}}{12})(x-\frac{6(1-\sqrt{1+28y(7y+1)})}{72})36(x−​12​​1+√​1+28y(7y+1)​​​​​)(x−​72​​6(1−√​1+28y(7y+1)​​​)​​)


7

 

Simplify \frac{6(1-\sqrt{1+28y(7y+1)})}{72}​72​​6(1−√​1+28y(7y+1)​​​)​​ to \frac{1-\sqrt{1+28y(7y+1)}}{12}​12​​1−√​1+28y(7y+1)​​​​​

36(x-\frac{1+\sqrt{1+28y(7y+1)}}{12})(x-\frac{1-\sqrt{1+28y(7y+1)}}{12})36(x−​12​​1+√​1+28y(7y+1)​​​​​)(x−​12​​1−√​1+28y(7y+1)​​​​​)


Done

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