(4y + 1)(2y +1) this should be the answer: 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
)
