garnentas3m

2021-12-30

To detrmine factor:

a)${r}^{2}-5r-14$

b)$3{m}^{2}+5m-2$

c)$6{p}^{2}+13pq-5{q}^{2}$

a)

b)

c)

ol3i4c5s4hr

Beginner2021-12-31Added 48 answers

Step 1

a) Given equation:${r}^{2}-5r-14$

To find integers b and d such that

${r}^{2}-5r-14=(r+b)(r+d)$

$={r}^{2}+dr+br+bd$

${r}^{2}-5r-14={r}^{2}+(b+d)r+bd$

Since:

must have$bd=-14$ , b and d are factors of -14

Similarly:$b+d=-5$

Table show the possibilities

$$\begin{array}{|cc|}\hline \text{Factors b, d of -14}& \text{Sum}\text{}b+d=-5\\ 2(-7)& 2-7=-5\\ 1(-14)& 1-14=-13\\ \hline\end{array}$$

So,${r}^{2}-5r-14=(r+2)(r-7)$

To check:

$(r+2)(r-7)=r(r-7)+2(r-7)$

$={r}^{2}-7r+2r-14$

$={r}^{2}-5r-14$

a) Given equation:

To find integers b and d such that

Since:

must have

Similarly:

Table show the possibilities

So,

To check:

Lynne Trussell

Beginner2022-01-01Added 32 answers

Step 1

b)$3{m}^{2}+5m-2$

To find integers

$3{m}^{2}+5m-2=(am+b)(cm+d)$

$=ac{m}^{2}+adm+bcm+bd$

$3{m}^{2}+5m-2=ac{m}^{2}+(ad+bc)m+bd$

Step 2

Since:

$ac=3$

$bd=-2$

$(3m-2)(m+1)=3m(m+1)-2(m+1)$

$=3{m}^{2}+3m-2m-2$

$=3{m}^{2}+m-2$

$(3m-1)(m+2)=3m(m+2)-1(m+2)$

$=3{m}^{2}+6m-m-2$

$=3{m}^{2}+5m-2$

Answer: The last trial gives the correct factorization.

b)

To find integers

Step 2

Since:

Answer: The last trial gives the correct factorization.

user_27qwe

Skilled2022-01-05Added 375 answers

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