How do you factor: $y= m^2-4m-21$ ?

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Answer 1 · 2015-12-10T13:41:20

$y= (m+3)(m-7)$

The things to note is that:
$(+3)xx(-7)=-21$
$-7+3=-4$

$color(blue)("Considering the RHS:")$

There are two condition to give $m^2$ and they are $(-m)xx(-m) and (+m)xx(+m)$

Lets try the $+m$

So we have: $(m+?)(m+?) " giving " m^2 + "something"$

Considering the numbers we look at first lets try:
$(m+3)(m-7) -> m^2-7m+3m-21$

But $-7m+3m = -4m$

$color(blue)("Putting it all together:")$

$y= (m+3)(m-7) =m^2-4m-21$

How do you factor: y= m^2-4m-21 y= m^2-4m-21 ?