Cem Hayes

2021-02-20

Decide the factorization of the polynomial ${m}^{2}+5m+mt+5t$

Yusuf Keller

Formula used:
The factors of a polynomial can be find by taking a common factor and this method is called factor by grouping,
$ab+ac+bd+cd=a\left(b+c\right)+d\left(b+c\right)$
$=\left(a+d\right)\left(b+c\right)$
Or,
$ab-ac+bd-cd=a\left(b-c\right)+d\left(b-c\right)$
$=\left(a+d\right)\left(b-c\right)$
Calculation:
Consider the polynomial ${m}^{2}+5m+mt+5t$.
This is a four term polynomial, factorization of this polynomial can be find by factor by grouping as,
${m}^{2}+5m+mt+5t=\left({m}^{2}+Sm\right)+\left(mt+5t\right)$
$=m\left(m+5\right)+t\left(m+5\right)$
As, $\left(m+5\right)$ is the common factor of the polynomial factor it out as follows:
${m}^{2}+5m+mt+5t=\left(m+5m\right)+\left(mt+5t\right)$
$=m\left(m+5\right)+t\left(m+5\right)$
$=\left(m+5\right)\left(m+1\right)$
The factorization of the polynomial ${m}^{2}+5m+mt+5t$ is $\left(m+5\right)\left(m+t\right)$.
Check the result as follows:
$\left(m+5\right)\left(m+t\right)=m\ast m+m\ast t+5\ast m+5\ast t$
$={m}^{2}+mt+5m+5t$
$={m}^{2}+5m+mt+5t$
Thus, the factorization of the polynomial ${m}^{2}+5m+mt+5t$ is $\left(m+5\right)\left(m+t\right)$.

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