Solving Quadratic Equations By Factoring Worksheet With Answer Key - Tallahassee Community College Page 4
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1
2
3
4
5x
12
0
x
1
0
−
=
+
=
12
12
1
1
+
+
−
−
x
12
x
1
=
=
−
The solutions are 12 and −1.
Example:
Solve: (y + 3)(y + 10) = −10
This example is similar to the first example we did, but with a big difference. The left hand side
of the equation is in factored form, but the equation is NOT equal to zero. So we cannot start
setting those two factors equal to zero at this point. First, we need to put the equation in standard
form by adding 10 to each side of the equation.
(
y
3
)(
y
10
)
10
+
+
=
−
10
10
+
+
(
y
3
)(
y
10
)
10
0
+
+
+
=
The 10 does not combine with the 10 or 3 in the parentheses. Now you will need to use the
FOIL method to multiply the two binomials.
y² + 10y + 3y + 30 + 10 = 0
Combine the like terms.
y² + 13y + 40 = 0
Now the equation is in standard form. The next step is to factor the polynomial.
(y + 8)(y + 5) = 0
Set each factor equal to 0 and solve each equation.
8
0
5
0
y
y
+
=
+
=
8
8
5
5
−
−
−
−
y
8
y
5
=
−
=
−
The solutions are −8 and −5.
Now you try these problems.
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