Question

What $\text{must be}\text{added to}(x^3+3x-8)\text{to get}$
$(3x^3+x^2+6)$ ?

Answer 1

Let the expression to be added be $Y$.
$(x^3+3x-8)+Y=(3x^3+x^2+6)$

$\text{To get}3x^3,\text{we need to add}2x^3\text{as}\text{first term to}{x}^3\text{in}(x^3+3x-8)$

$\text{Similarly, as we don't have}{x}^{2}in(x^3+3x-8).\text{We need to add}{x}^2\text{as second term}$

$\text{Similarly, we have an extra 3x in}(x^3+3x-8).\text{So, we need to add -3x.}\text{as third term}$

$\text{Similarly, to make -8 to +6 we need}\text{add 6 + 8 = 14 to}(x^3+3x-8).\text{So, we need to add 14}\text{as fourth term}$

$\text{So we have to add}(2x^3+x^2-3x+14)to(x^3+3x-8),\text{to get the expression}(3x^3+x^2+6).$