garnentas3m

2021-12-30

To detrmine factor:
a) ${r}^{2}-5r-14$
b) $3{m}^{2}+5m-2$
c) $6{p}^{2}+13pq-5{q}^{2}$

ol3i4c5s4hr

Step 1
a) Given equation: ${r}^{2}-5r-14$
To find integers b and d such that
${r}^{2}-5r-14=\left(r+b\right)\left(r+d\right)$
$={r}^{2}+dr+br+bd$
${r}^{2}-5r-14={r}^{2}+\left(b+d\right)r+bd$
Since:
must have $bd=-14$, b and d are factors of -14
Similarly: $b+d=-5$
Table show the possibilities

So, ${r}^{2}-5r-14=\left(r+2\right)\left(r-7\right)$
To check:
$\left(r+2\right)\left(r-7\right)=r\left(r-7\right)+2\left(r-7\right)$
$={r}^{2}-7r+2r-14$
$={r}^{2}-5r-14$

Lynne Trussell

Step 1
b) $3{m}^{2}+5m-2$
To find integers
$3{m}^{2}+5m-2=\left(am+b\right)\left(cm+d\right)$
$=ac{m}^{2}+adm+bcm+bd$
$3{m}^{2}+5m-2=ac{m}^{2}+\left(ad+bc\right)m+bd$
Step 2
Since:
$ac=3$
$bd=-2$
$\left(3m-2\right)\left(m+1\right)=3m\left(m+1\right)-2\left(m+1\right)$
$=3{m}^{2}+3m-2m-2$
$=3{m}^{2}+m-2$
$\left(3m-1\right)\left(m+2\right)=3m\left(m+2\right)-1\left(m+2\right)$
$=3{m}^{2}+6m-m-2$
$=3{m}^{2}+5m-2$
Answer: The last trial gives the correct factorization.

user_27qwe