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Let’s check out your problem:
Re-write the quadratic function below in Standard Form
\newline
y
=
7
(
x
+
9
)
(
x
+
1
)
y=7(x+9)(x+1)
y
=
7
(
x
+
9
)
(
x
+
1
)
\newline
Answer:
y
=
y=
y
=
View step-by-step help
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Math Problems
Algebra 1
Multiply two binomials
Full solution
Q.
Re-write the quadratic function below in Standard Form
\newline
y
=
7
(
x
+
9
)
(
x
+
1
)
y=7(x+9)(x+1)
y
=
7
(
x
+
9
)
(
x
+
1
)
\newline
Answer:
y
=
y=
y
=
Combine like terms:
Now, we will simplify the expression by combining like terms.
\newline
y
=
7
[
x
2
+
x
+
9
x
+
9
]
y = 7[x^2 + x + 9x + 9]
y
=
7
[
x
2
+
x
+
9
x
+
9
]
\newline
y
=
7
[
x
2
+
10
x
+
9
]
y = 7[x^2 + 10x + 9]
y
=
7
[
x
2
+
10
x
+
9
]
Distribute
7
7
7
:
Next, we distribute the
7
7
7
across each term inside the brackets to get the standard form of the
quadratic equation
.
\newline
y
=
7
x
2
+
70
x
+
63
y = 7x^2 + 70x + 63
y
=
7
x
2
+
70
x
+
63
Standard form:
The quadratic function is now in standard form, which is
y
=
a
x
2
+
b
x
+
c
y = ax^2 + bx + c
y
=
a
x
2
+
b
x
+
c
, where
a
a
a
,
b
b
b
, and
c
c
c
are constants.
\newline
In this case,
a
=
7
a = 7
a
=
7
,
b
=
70
b = 70
b
=
70
, and
c
=
63
c = 63
c
=
63
.
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\newline
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−
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\newline
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\text{[[even][odd][neither]]}
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\newline
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)
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r
+
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)
\newline
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Question
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\newline
(
x
+
7
)
(
x
+
4
)
(x + 7)(x + 4)
(
x
+
7
)
(
x
+
4
)
\newline
Write your answer as a list of values separated by commas.
\newline
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=
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