This is the last question PLSSSSS HELP
2 answers:
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Answer:
The answer is in pictures.
hope it helps:)
Most real number arithmetic is pretty vacuous
That's about as good a way as possible to write this particular real number. But it's far from the only way.
so
Sometimes you can factor something out and you have a common radical:
But most of them are vacuous,
Answer:
Step-by-step explanation:
(3a + 2b − c) (a − b + 2c)
Expand (3a + 2b − c) (a − b + 2c) by multiplying each term in the first expression by each term in the second expression.
3a ⋅ a + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)
3 (a ⋅ a) + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)
3a^2 + 3a (−b) + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)
Rewrite using the commutative property of multiplication.
3a^2 + 3 ⋅ −1ab + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)
3a^2 − 3ab + 3a (2c) + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)
Rewrite using the commutative property of multiplication.
3a^2 − 3ab + 3 ⋅ 2ac + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)
3a^2 − 3ab + 6ac + 2ba + 2b (−b) + 2b (2c) − ca − c (−b) − c (2c)
Rewrite using the commutative property of multiplication.
3a^2 − 3ab + 6ac + 2ba + 2 ⋅ −1b ⋅ b + 2b (2c) − ca − c (−b) − c (2c)
3a^2 − 3ab + 6ac + 2ba + 2 ⋅ −1 (b ⋅ b) + 2b (2c) − ca − c (−b) − c (2c)
3a^2 − 3ab + 6ac + 2ba + 2 ⋅ −1b^2 + 2b (2c) − ca − c (−b) − c (2c)
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 2b (2c) − ca − c (−b) − c (2c)
Rewrite using the commutative property of multiplication.
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 2 ⋅ 2bc − ca − c (−b) − c (2c)
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca − c (−b) − c (2c)
Rewrite using the commutative property of multiplication.
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca − 1 ⋅ −1cb − c (2c)
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + 1cb − c (2c)
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − c (2c)
Rewrite using the commutative property of multiplication.
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 1 ⋅ 2c ⋅ c
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 1 ⋅ 2 (c ⋅ c)
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 1 ⋅ 2c^2
3a^2 − 3ab + 6ac + 2ba − 2b^2 + 4bc − ca + cb − 2c^2
3a^2 − 3ab + 2ab + 6ac − 2b^2 + 4bc − ca + cb − 2c^2
3a^2 − ab + 6ac − 2b^2 + 4bc − ca + cb − 2c^2
3a^2 − ab − 2b^2 + 4bc + 6ac − 1ac + cb − 2c^2
3a^2 − ab − 2b^2 + 4bc + 5ac + cb − 2c^2
3a^2 − ab − 2b^2 + 4bc + bc + 5ac − 2c^2
3a^2 − ab − 2b^2 + 5bc + 5ac − 2c^2
3a^2 − ab − 2b^2 + 5ac + 5bc − 2c^2
3a^2 − ab + 5ac − 2b^2 + 5bc − 2c^2