Is it possible to factor $y=x^2-4x-3$ ? If so, what are the factors?

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Answer 1 · 2017-02-16T16:50:28

Factors are $y=(x-2+sqrt7)(x-2-sqrt7)$

When you generally say - Ïs it possible to factor $y=ax^2+bx+c$ , what is normally meant is to have rational factors of type $y=a(x+p)(x+q)$

This is possible if the discriminant $b^2-4ac$ is square of a rational number.

Here in $y=x^2-4x-3$ , we have $a=1$ , $b=-4$ and $c=-3$ and hence $b^2-4ac=(-4)^2-4xx1xx(-3)=16+12=28$

as it is not the square of a rational number you cannot have rational factors. But it is still possible to have irrational factors.

As $y=x^2-4x-3=(x^2-2xx2xx x+2^2)-2^2-3$

= $(x-2)^2-7=(x-2)^2-(sqrt7)^2$

and as $a^2-b^2=(a+b)(a-b)$

factors are $y=(x-2+sqrt7)(x-2-sqrt7)$ .

Is it possible to factor y=x^2-4x-3 y=x^2-4x-3 ? If so, what are the factors?