Question

Factoring quadratics. Not in a rush, but I would appreciate some help on this. It's too late into the day to ask for help from my teacher. Explain how you did it step by step in the answer. There are more bots now, so please try to be helpful for once..

Answer 1

Answer:

• (y + 10)(2y + 1)

Step-by-step explanation:

Factor the expression:

• 2y² + 21y + 10 =
• 2y² + 20y + y + 10 =
• 2y(y + 10) + (y + 10) =
• (y + 10)(2y + 1)

Answer 2

Answer:

(y+10)(2y+1)

Step-by-step explanation:

2y^2 +21y+10

Factor the expression by grouping. First, the expression needs to be rewritten as 2y^2 +ay+by+10. To find a and b, set up a system to be solved.

a+b=21

ab=2×10=20

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 20.

1,20

2,10

4,5

Calculate the sum for each pair.

1+20=21

2+10=12

4+5=9

The solution is the pair that gives sum 21.

a=1

b=20

Rewrite 2y^2 +21y+10 as (2y^2 +y)+(20y+10).

(2y ^2 +y)+(20y+10)

actor out y in the first and 10 in the second group.

y(2y+1)+10(2y+1)

Factor out common term 2y+1 by using distributive property.

(2y+1)(y+10)