(y ^3+ 2y^2+3y+9) ÷ (y - 1) by long divison
Answers
Answer:
know the question asked for long division, but there is an easier way to do it.
We want
2
y
2
−
3
y
+
1
y
−
2
First, let
f
(
y
)
=
2
y
2
−
3
y
+
1
Then take
f
(
2
)
=
2
(
2
2
)
−
3
(
2
)
+
1
=
3
So
2
y
2
−
3
y
+
1
y
−
2
has a remainder of
3
.
Therefore
2
y
2
−
3
y
+
1
=
(
a
y
+
b
)
(
y
−
2
)
+
3
2
y
2
−
3
y
−
2
=
(
a
y
+
b
)
(
y
−
2
)
y
2
has a coefficient of
2
, so
a
y
2
=
2
y
2
⇒
a
=
2
The constant is
−
2
, so
−
2
b
=
−
2
⇒
b
=
1
Thus the quotient,
(
a
y
+
b
)
, equals
(
2
y
+
1
)
Answer link
Answer:
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Step-by-step explanation: