$(x + 4)^{2} - 5 x y - 20 y - 6 y^{2} =$

(x−6y+4)(x+y+4)

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(x+6y+4)(x+y+4)

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(x−6y+4)(x−y−4)

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(x−6y+4)(x−y+4)

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Solution

The correct option is A
(x−6y+4)(x+y+4)

$(x + 4)^{2} - 5 x y - 20 y - 6 y^{2}$

$(x + 4)^{2} - 5 y (x + 4) - 6 y^{2}$

$(x + 4)^{2} + y (x + 4) - 6 y (x + 4) - 6 y^{2}$

$(x + 4) [x + 4 + y] - 6 y [x + 4 + y]$

$[x + 4 - 6 y] [x + 4 + y]$

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Similar questions

\frac{x}{y}

x^{2} + 6 y^{2} = 5 x y

Find $\frac{x}{y}$ ; when $x^{2} + 6 y^{2} = 5 x y$ .

6 x^{2} - 5 x y - 6 y^{2} = 0

6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0

6 x^{2} - 5 x y - 6 y^{2} = 0

6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0

x^{2} - 4 x y + 6 y^{2} - 2 x - 20 y = 29

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