Find N if N - (30u^2 v - 15u v^2 + v^3) = 12u^2 v + 2v^3 - 10u v^2.

Step-by-step solution

1. 1

   Isolate $N$ by adding the subtracted polynomial to both sides. Treat the whole trinomial as one object rather than distributing first:

   $N=\left(12u^2v+2v^3-10uv^2\right)+\left(30u^2v-15uv^2+v^3\right)$

   This is safer than moving terms individually, because the minus sign in front of the bracket has already been dealt with by the addition.

2. 2

   Sort both polynomials into matching term types. Writing each in the order $u^2v$, $uv^2$, $v^3$:

   $12u^2v-10uv^2+2v^3\qquad\text{and}\qquad 30u^2v-15uv^2+v^3$

   Note that $u^2v$ and $uv^2$ are not like terms — they differ in which variable is squared, and confusing them is the standard error here.

3. 3

   Add the coefficients of each type.

   $u^2v:\;12+30=42,\qquad uv^2:\;-10+(-15)=-25,\qquad v^3:\;2+1=3$

4. 4

   Write the answer.

   $N=42u^2v-25uv^2+3v^3$

5. 5

   Check by substituting back into the original equation. Take $u=1$, $v=2$: then $N=42(2)-25(4)+3(8)=84-100+24=8$. The subtracted trinomial is $30(2)-15(4)+8=60-60+8=8$, so the left side is $8-8=0$. The right side is $12(2)+2(8)-10(4)=24+16-40=0\;\checkmark$. A second test at $u=2$, $v=1$ gives $N=168-50+3=121$, subtracted trinomial $120-30+1=91$, left side $30$; right side $48+2-20=30\;\checkmark$.