16y^2-x^2-6x-9. How can I factorise this? I know that -(x^2+6x+9) is equal to -(X+3)^2, and 16y^2 can be written as (4y)^2 but I can't quite get the solution (4y-x-3)(4y+X+3)

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Answer 1 · 2018-02-28T08:37:12

$16y^2-x^2-6x-9=(4y-x-3)(4y+x+3)$

First observe that $16y^2-x^2-6x-9$

= $16y^2-(x^2+6x+9)$

= $(4y)^2-(x+3)^2$

Now observe that this is of type $A^2-B^2$ , where $A=4y$ and $B=(x+3)$ . Now factors of $A^2-B^2=(A-B)(A+B)$

Hence we can write $(4y)^2-(x+3)^2$ as

$(4y-(x+3))(4y+(x+3))$

= $(4y-x-3)(4y+x+3)$