Mixed Review Of Factoring Worksheet Page 2
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1
2
32
2
5.
9m
– 4z
is a difference of squares, so we get:
2
2
2
2
6(9m
– 4z
)
F
– L
= (F – L)(F + L); F = 3m, L = 2z
= 6(3m – 2z)(3m + 2z)
6.
Everything is factored, so the answer is 6(3m – 2z)(3m + 2z).
5
3
2
c)
x
– 36x
– x
+ 36
1.
The G.C.F. is 1.
2.
The polynomial inside the parentheses has 4 terms.
3.
We will group the first two and the last two terms:
5
3
2
5
3
2
3
x
– 36x
– x
+ 36 = (x
– 36x
) + (– x
+ 36) (factor out x
& 1)
3
2
2
2
= x
(x
– 36) – 1(x
– 36)
(factor out x
– 36)
2
3
= (x
– 36)(x
– 1)
6.
We can break down both factors:
2
5) x
– 36 has two terms and it is a difference of squares:
2
2
2
x
– 36 F
– L
= (F – L)(F + L); F = x, L = 6
= (x – 6)(x + 6)
3
5) x
– 1 has two terms and it is a difference of cubes:
3
3
3
2
2
x
– 1
F
– L
= (F – L)(F
+ FL + L
); F = x, L = 1
2
= (x – 1)(x
+ x + 1)
2
3
2
Thus, (x
– 36)(x
– 1) = (x – 6)(x + 6)(x – 1)(x
+ x + 1)
2
6) Since x
+ x + 1 has degree of 2, it is prime. Thus,
everything is completely factored.
2
Our answer is (x – 6)(x + 6)(x – 1)(x
+ x + 1).
4
4
d)
3000a
b + 81ab
1.
The G.C.F. is 3ab. Factoring out 3ab, we get:
4
4
3000a
b + 81ab
3
3
= 3ab(1000a
+ 27b
)
2.
The polynomial inside the parentheses has 2 terms.
3
3
5.
1000a
+ 27b
is not a difference of squares, but is a sum of
3
3
cubes:
1000a
+ 27b
3
3
3
3
2
2
= (10a)
+ (3b)
F
+ L
= (F + L)(F
– FL + L
)
2
2
= (10a + 3b)(100a
– 30ab + 9b
)
3
3
Thus, 3ab(1000a
+ 27b
)
2
2
= 3ab(10a + 3b)(100a
– 30ab + 9b
)
2
2
6.
Since 100a
– 30ab + 9b
is degree 2, it is prime. Everything is
completely factored.
2
2
Our answer is 3ab(10a + 3b)(100a
– 30ab + 9b
).
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