Q8. - Algebra 1

Question

Factor the polynomial $y^2+\mathbf{9}y+\mathbf{20}$.

$$\begin{gathered} \mathbf{F}\left(y+\mathbf{2}\right)\left(y+\mathbf{10}\right) \\ \mathbf{G}\left(y+\mathbf{4}\right)\left(y+\mathbf{5}\right) \\ \mathbf{H}\left(y+\mathbf{3}\right)\left(y+\mathbf{8}\right) \\ \mathbf{J}\left(y+\mathbf{5}\right)\left(y+\mathbf{2}\right) \end{gathered}$$

Step-by-Step Solution

Answer

Factors of $y^2+9y+20$ are $\left(y+\mathbf{4}\right)\left(y+\mathbf{5}\right)$.

Step 1. Compare.

Compare a given trinomial to $a{x}^2+bx+c$ and write the values of a,b and c.

$$\begin{gathered} a=1 \\ b=9 \\ c=20 \end{gathered}$$

Step 2. List the factors.

List the factors of 20. Look for the pair of factors with a sum of 9.

 

Factors of	Sum of factors
1,20	21
2,10	12
4,5	9

 

Therefore, correct factors are 4 and 5.

Step 3. Apply factoring pattern of a x 2 + b x + c .

There are two numbers having product of 20 and sum of 9. The numbers are 4 and 5.

$$\begin{gathered} y^2+9y+20=\left(y+m\right)\left(y+p\right) \\ =\left(y+4\right)\left(y+5\right) \end{gathered}$$

 

Therefore, the factors of $y^2+9y+20$ are $\left(y+4\right)\left(y+5\right)$.

 

The obtained solution completely matches with option G. Therefore, option G is the required answer.