[tex]\large\sf\underline{Given\::}[/tex] [tex]\sf\:\frac{m^{2}-14m-32}{(m+2) }[/tex] [tex]\large\sf\underline{To\::}[/tex] Divide the given expression [tex]\large\sf\underline{Solution\::}[/tex] [tex]\sf\:\frac{m^{2}-14m-32}{(m+2) }[/tex] Let’s factorise the numerator by middle term breaking So for factorising we need such two terms whose product gives us 32 and whose sum or difference would give us middle term ( 14 ) . That two terms would be 16 and 2 . 16 × 2 = 32 16 – 2 = 14 Now substituting the two terms in the expression : [tex]\sf\longrightarrow\:\frac{m^{2}-(16-2)m-32}{(m+2) }[/tex] Multiplying the terms in numerator [tex]\sf\longrightarrow\:\frac{m^{2}-16m+2m-32}{(m+2) }[/tex] Taking m from first two terms and 2 from second two terms as common in numerator [tex]\sf\longrightarrow\:\frac{m(m-16)+2(m-16)}{(m+2) }[/tex] Taking ( m – 16 ) common from whole terms in numerator [tex]\sf\longrightarrow\:\frac{(m-16)(m+2)}{(m+2) }[/tex] [tex]\sf\longrightarrow\:\frac{(m-16)\cancel{(m+2)}}{\cancel{(m+2)}}[/tex] [tex]\small\fbox\red{★\:(\:m\:-\:16\:)}[/tex] !! Hope it helps !! Reply
[tex]\large\sf\underline{Given\::}[/tex]
[tex]\large\sf\underline{To\::}[/tex]
[tex]\large\sf\underline{Solution\::}[/tex]
[tex]\sf\:\frac{m^{2}-14m-32}{(m+2) }[/tex]
So for factorising we need such two terms whose product gives us 32 and whose sum or difference would give us middle term ( 14 ) .
That two terms would be 16 and 2 .
Now substituting the two terms in the expression :
[tex]\sf\longrightarrow\:\frac{m^{2}-(16-2)m-32}{(m+2) }[/tex]
[tex]\sf\longrightarrow\:\frac{m^{2}-16m+2m-32}{(m+2) }[/tex]
[tex]\sf\longrightarrow\:\frac{m(m-16)+2(m-16)}{(m+2) }[/tex]
[tex]\sf\longrightarrow\:\frac{(m-16)(m+2)}{(m+2) }[/tex]
[tex]\sf\longrightarrow\:\frac{(m-16)\cancel{(m+2)}}{\cancel{(m+2)}}[/tex]
[tex]\small\fbox\red{★\:(\:m\:-\:16\:)}[/tex]
!! Hope it helps !!