Question

Find the sum of 7a^3 + 14a +12 and -6a^3 +12a^2 -7

Answer 1

To find the sum:


`{7a}^(3) + 14a + 12 + ( {- 6a }^(3) ) + {12a }^(2) - 7 \\ = {7a }^(3) - {6a}^(3) + {12a }^(2) + 14a + 12 - 7 \\ = {a}^(3) + {12a }^(2) + 14a + 5`
Therefore the answer is a^3+12a^2+14a+5.
Hope it helps!

Answer 2

`a^3+12\cdot{a^2}+14\cdot{a}+5`

Explanation:

We can write the sum of the two expression as:

`7\cdot{a^3}+14\cdot{a}+12+\cdot{(-6\cdot{a^3}+12\cdot{a^2}-7)}`

We can multiply the plus sign in the second expression:

`7\cdot{a^3}+14\cdot{a}+12-6\cdot{a^3}+12\cdot{a^2}-7`

We must group the terms of a³, a², a terms and normal numbers:

`(7-6)\cdot{a^3}+14\cdot{a}+12\cdot{a^2}-7+12`

`a^3+14\cdot{a}+12\cdot{a^2}+5`

`a^3+12\cdot{a^2}+14\cdot{a}+5`