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

Answer:

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

Step-by-step 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:

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

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

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