Question

factorise 6x^2+35xy-6y^2

Answer 1

6x^2+35xy-6y^2 = 6x^2+36xy-1.xy-6y^2
= 6x(x+6y) -y(x+6y)
=(x+6y)(6x-y)
therefore 6x^2+35xy-6y^2 = (x+6y)(6x-y)

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Answer 2

Method 1.

Now, 6x² + 35xy - 6y²

= 6x² + (36 - 1) xy - 6y²

= 6x² + 36xy - xy - 6y²

= 6x (x + 6y) - y (x + 6y)

= (x + 6y) (6x - y)

This is the required factorization.

Method 2.

We can use a different method to find the factorization, only when you can solve quadratic equations using Sridhar Acharya's formula.

We equate the given expression with zero (0).

Now, 6x² + 35xy - 6y² = 0

or, (6) x² + (35y) x + (- 6y²) = 0

Using quadratic formula, we get

x = (- 35y ± √[(35y)² - {4 * 6 * (- 6y²)}] )/(2 * 6)

= {- 35y ± √(1369y²}/12

= (- 35y ± 37y)/12

or, x = 2y/12, - 72y/12

So we can form an equation whose roots are

x = 2y/12 and x = - 72y/12

Thus the required equation is

(x - 2y/12) {x - (- 72y)/12} = 0

or, (x - y/6) (x + 6y) = 0

or, (6x - y) (x + 6y) = 0

Now exclude the = 0 from the right hand side, and we get the required factorization as

(6x - y) (x + 6y).