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Teacher: Mr. Joseph
Factoring Trinomials Review
Name:__________________________
Period: ________ Date: ____________
Factor:
1)
25r2 + 5r -6
2)
12h2 + h - 6
3)
3b2 + 10b + 8
4)
5h2 + 8h + 3
5)
6b2 + b -2
6)
10v2 + 9v + 2
7)
s4 - 16e4
8)
16x2 -9p4
9)
x2 -9z4
10)
u2 -c4
11)
s4 - 81g4
12)
x8 -16x6
13)
9n2 -30n + 25
14)
16w2 + 40w + 25
15)
d2 + 4d + 4
16)
3h8 -12h7 + 12h6
17)
4m2 -4m + 1
18)
v2 + 8v + 16
EasyWorksheet Step By Step Answers
User Name: J. Joseph
Form #469414141957
Practice Problems
1) 25r2 + 5r -6
First we fill in the first term in the top left corner and the last term in the bottom right:
??r ??
??r 25r2
-6
??
Nert use the factors of 25 to fill in the values by r, and then use the factors of -6 to fill
in the values represented by ??'s.
Then, we try putting different values up on top and on the side until we have what
looks like a normal multiplication table which works out to the answer.
Notice that the diagonal elements add to our middle term!
5r
-2
2
5r 25r -10r
3 15r -6
So 25r2 + 5r -6 factors into (5r -2)(5r + 3)
2) 12h2 + h -6
First we fill in the first term in the top left corner and the last term in the bottom right:
??h ??
??h 12h2
??
-6
Neht use the factors of 12 to fill in the values by h, and then use the factors of -6 to fill
in the values represented by ??'s.
Then, we try putting different values up on top and on the side until we have what
looks like a normal multiplication table which works out to the answer.
Notice that the diagonal elements add to our middle term!
3h
-2
4h 12h2 -8h
3 9h -6
So 12h2 + h -6 factors into (3h -2)(4h + 3)
3) 3b2 + 10b + 8
First we fill in the first term in the top left corner and the last term in the bottom right:
??b ??
??b 3b2
8
??
Nebt use the factors of 3 to fill in the values by b, and then use the factors of 8 to fill
in the values represented by ??'s.
Then, we try putting different values up on top and on the side until we have what
looks like a normal multiplication table which works out to the answer.
Notice that the diagonal elements add to our middle term!
b
2
3b 3b2 6b
4 4b 8
So 3b2 + 10b + 8 factors into (b + 2)(3b + 4)
4) 5h2 + 8h + 3
First we fill in the first term in the top left corner and the last term in the bottom right:
??h ??
??h 5h2
??
3
Neht use the factors of 5 to fill in the values by h, and then use the factors of 3 to fill
in the values represented by ??'s.
Then, we try putting different values up on top and on the side until we have what
looks like a normal multiplication table which works out to the answer.
Notice that the diagonal elements add to our middle term!
5h 3
h 5h2 3h
1 5h 3
So 5h2 + 8h + 3 factors into (5h + 3)(h + 1)
5) 6b2 + b -2
First we fill in the first term in the top left corner and the last term in the bottom right:
??b ??
??b 6b2
-2
??
Nebt use the factors of 6 to fill in the values by b, and then use the factors of -2 to fill
in the values represented by ??'s.
Then, we try putting different values up on top and on the side until we have what
looks like a normal multiplication table which works out to the answer.
Notice that the diagonal elements add to our middle term!
3b 2
2b 6b2 4b
-1 -3b -2
So 6b2 + b -2 factors into (3b + 2)(2b -1)
6) 10v2 + 9v + 2
First we fill in the first term in the top left corner and the last term in the bottom right:
??v ??
??v 10v2
??
2
Nevt use the factors of 10 to fill in the values by v, and then use the factors of 2 to fill
in the values represented by ??'s.
Then, we try putting different values up on top and on the side until we have what
looks like a normal multiplication table which works out to the answer.
Notice that the diagonal elements add to our middle term!
5v
2
2
2v 10v 4v
1 5v 2
So 10v2 + 9v + 2 factors into (5v + 2)(2v + 1)
7) s4 - 16e4
Since s2 is the square root of s4, and 4e2 is the square root of 16e4,
we simply write them once with a + between them and once with a - between them to
get:
s4 - 16e4 factors into (s2 + 4e2)(s2 - 4e2)
Now, since s is the square root of s2, and 2e is the square root of 4e2,
we simply write them once with a + between them and once with a - between them to
get:
And we find s2 - 4e2 factors into (s + 2e)(s - 2e)
So you end up with: (s + 2e)(s - 2e)(s2 + 4e2)
8) 16x2 -9p4
Since 4x is the square root of 16x2, and 3p2 is the square root of 9p4,
we simply write them once with a + between them and once with a - between them to
get:
16x2 -9p4 factors into (4x + 3p2)(4x -3p2)
9) x2 -9z4
Since x is the square root of x2, and 3z2 is the square root of 9z4,
we simply write them once with a + between them and once with a - between them to
get:
x2 -9z4 factors into (x + 3z2)(x -3z2)
10) u2 -c4
Since u is the square root of u2, and c2 is the square root of c4,
we simply write them once with a + between them and once with a - between them to
get:
u2 -c4 factors into (u +c2)(u -c2)
11) s4 - 81g4
Since s2 is the square root of s4, and 9g2 is the square root of 81g4,
we simply write them once with a + between them and once with a - between them to
get:
s4 - 81g4 factors into (s2 + 9g2)(s2 - 9g2)
Now, since s is the square root of s2, and 3g is the square root of 9g2,
we simply write them once with a + between them and once with a - between them to
get:
And we find s2 - 9g2 factors into (s + 3g)(s - 3g)
So you end up with: (s + 3g)(s - 3g)(s2 + 9g2)
12) x8 -16x6
You can pull out x6 to get: x2 -16
Since x is the square root of x2, and 4 is the square root of 16,
we simply write them once with a + between them and once with a - between them to
get:
x2 -16 factors into (x + 4)(x -4)
So you end up with:(x6)(x + 4)(x -4)
13) 9n2 -30n + 25
Notice that the first coefficient and the last are both square numbers! If we multiply
these coefficients together, we get 225 which is the square of -15, our middle term
divided by 2!
So take the square root of the first term, 3, and the square root of the last term, 5. Then
put them together with the sign of the middle term, and we get: (3n -5)2 as our answer.
14) 16w2 + 40w + 25
Notice that the first coefficient and the last are both square numbers! If we multiply
these coefficients together, we get 400 which is the square of 20, our middle term
divided by 2!
So take the square root of the first term, 4, and the square root of the last term, 5. Then
put them together with the sign of the middle term, and we get: (4w + 5)2 as our
answer.
15) d2 + 4d + 4
Notice that the first coefficient and the last are both square numbers! If we multiply
these coefficients together, we get 4 which is the square of 2, our middle term divided
by 2!
So take the square root of the first term, 1, and the square root of the last term, 2. Then
put them together with the sign of the middle term, and we get: (d + 2)2 as our answer.
16) 3h8 -12h7 + 12h6
You can pull out 3h6 to get: h2 -4h + 4
Use a BOX or other factoring technique to factor h2 -4h + 4 into (h -2)2
So you end up with:(3h6)(h -2)2
17) 4m2 -4m + 1
Notice that the first coefficient and the last are both square numbers! If we multiply
these coefficients together, we get 4 which is the square of -2, our middle term
divided by 2!
So take the square root of the first term, 2, and the square root of the last term, 1. Then
put them together with the sign of the middle term, and we get: (2m -1)2 as our
answer.
18) v2 + 8v + 16
Notice that the first coefficient and the last are both square numbers! If we multiply
these coefficients together, we get 16 which is the square of 4, our middle term
divided by 2!
So take the square root of the first term, 1, and the square root of the last term, 4. Then
put them together with the sign of the middle term, and we get: (v + 4)2 as our
answer.