Question

Add 5t⁴+2t+12, -4t²+t, 2t⁴-2t²-2​

Answer 1

From the given question the correct answer is :

The addition of 5t⁴+2t+12, -4t²+t, 2t⁴-2t²-2​ is 6t⁴-6t²+ 3t+10

Given :  5t⁴+2t+12, -4t²+t, 2t⁴-2t²-2​

To find :

addition of the equation

Solution:

5t⁴+2t+12 + ,-4t²+t, +2t⁴-2t²-2​

From the rule of inter we can add this

= 5t⁴+2t+12 -4t²+t +2t⁴-2t²-2​

First we will add the number having same variable

so,

=5t⁴+2t+12 -4t²+t +2t⁴-2t²-2​

=6t⁴ + 2t+12 -4t²+t-2t²-2​

=6t⁴ + 3t+12-4t²-2t²-2

= 6t⁴ + 3t+10-4t²-2t²

=6t⁴ + 3t+10-6t²

= now arrange the equation with respect to its power

=6t⁴-6t²+ 3t+10

hence, The addition of 5t⁴+2t+12, -4t²+t, 2t⁴-2t²-2​ is 6t⁴-6t²+ 3t+10

Answer 2

❍ Given :-

• 5t⁴ + 2t + 12
• -4t² + t
• 2t⁴ - 2t² - 2

❍ To Find :-

• Sum of all terms given

❍ Solution :-

$\sf \small → {5t}^{2} + 2t + 12 + ( - {4t}^{2} + t) + ( {2t}^{4} - {2t}^{2} - 2)$

➸ Opening of the brackets

$\sf \small → {5t}^{2} + 2t + 12 - {4t}^{2} + t + {2t}^{4} - {2t}^{2} - 2$

➸ Rearranging terms,

$\sf \small → {5t}^{4} + {2t}^{4} - {4t}^{2} - {2t}^{2} + 2t + t + 12 - 2$

$\sf \small → {7t}^{4} - {6t}^{2} + {2t} + 10$

Hence, the sum of given terms is 7t⁴ - 6t² + 2t + 10.