David Young

2021-12-28

Identify the factor:
a) ${r}^{2}+7r+10$
b) ${x}^{2}+4x+3$
c) ${y}^{2}+6y+8$

Rita Miller

Step 1
a) We can get a quadratic by multiplying two first-degree polynomials.
Method:
$\left(x+4\right)\left(x-2\right)=x\left(x-2\right)+4\left(x-2\right)$
$={x}^{2}-2x+4x-8$
$={x}^{2}+2x-8$
Step 2
Given:
${r}^{2}+7r+10$
We have to find integers b and d such that
${r}^{2}+7r+10=\left(r+b\right)\left(r+d\right)$
$={r}^{2}+dr+br+bd$
${r}^{2}+7r+10={r}^{2}+\left(b+d\right)r+bd$
Since the constant coefficients on each side of the equation ought to be equal, we must have $bd=10$ (i.e) b and d are factors of 10.
Similarly, the coefficients of r must be the same, so that $b+d=7$
The following table shows the possibilities.
$\begin{array}{|cc|}\hline \text{Factors b, d of 10}& \text{Sum}b+d=7\\ 5×2& 5+2=7\\ \hline\end{array}$
There is no need to list negative factors, such as $\left(-5\right)\left(-2\right)$, since their sum is negative. So the factors are 5 and 2.
To check:
$\left(r+5\right)\left(r+2\right)=r\left(r+2\right)+5\left(r+2\right)$
${r}^{2}+2r+5r+10$
$={r}^{2}+7r+10$

amarantha41

Step 1
Given factor:
${x}^{2}+4x+3$
${x}^{2}+4x+3=\left(x+b\right)\left(x+d\right)$
$={x}^{2}+dx+bx+bd$
${x}^{2}+4x+3={x}^{2}+\left(b+d\right)x+bd$
Following table
$\begin{array}{|cc|}\hline \text{Factors b, d of 3}& \text{Sum}b+d=4\\ 3×1& 3+1=4\\ \hline\end{array}$
To check:
$\left(x+3\right)\left(x+1\right)=x\left(x+1\right)+3\left(x+1\right)$
$={x}^{2}+x+3x+3$
$={x}^{2}+4x+3$

user_27qwe

$\begin{array}{}\\ \text{Step 1}\\ \text{Given:}{y}^{2}+6y+8\\ \text{Find integers b and d such that}\\ {y}^{2}+6y+8=\left(y+b\right)\left(y+d\right)\\ ={y}^{2}+dy+by+bd\\ {y}^{2}+6y+8={y}^{2}+\left(B+d\right)y+bd\\ \text{Step 2}\\ \text{Table:}\\ \begin{array}{|cc|}\hline \text{Factors b, d of 8}& \text{Sum}b+d=6\\ 4×2& 4+2=6\\ 1×8& 1+8=9\\ \hline\end{array}\\ \text{Step 3}\\ \text{To check:}\\ \left(y+4\right)\left(y+2\right)=y\left(y+2\right)+4\left(y+2\right)\\ ={y}^{2}+2y+4y+8\\ ={y}^{2}+6y+8\end{array}$