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

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Answer 1 · 2015-12-10T13:41:20

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

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

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

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

Lets try the $+ m$ $+m$

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

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

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

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

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

How do you factor: y= m^2-4m-21 y=m2−4m−21y= m^2-4m-21 ?