Factor completely

8y²+6y+1

(2y +1)(4y + 1) is the factored form

(4y + 1)(2y +1) this should be the How to solve your problem
8

2
+
6

+
1
8
y
2
+
6
y
+
1
Grouping
1
Use the sum-product pattern
8

2
+
6

+
1
8
y
2
+
6
y
+
1
8

2
+
4

+
2

+
1
8
y
2
+
4
y
+
2
y
+
1
2
Common factor from the two pairs
8

2
+
4

+
2

+
1
8
y
2
+
4
y
+
2
y
+
1
4

(
2

+
1
)
+
2

+
1
4
y
(
2
y
+
1
)
+
2
y
+
1
3
Rewrite in factored form
4

(
2

+
1
)
+
2

+
1
4
y
(
2
y
+
1
)
+
2
y
+
1
(
4

+
1
)
(
2

+
1
)
(
4
y
+
1
)
(
2
y
+
1
)
Solution
(
4

+
1
)
(
2

+
1
)