Find the product (-2x^2 + 7)(9x^2 - 4x + 2), writing the multiplication horizontally.

Step-by-step solution

1. 1

   Note what makes this one easier. The first factor $-2x^2+7$ has no $x$ term, so the six partial products land in only five degree slots and just one slot ends up with two contributions. Keeping the minus sign glued to $-2x^2$ is the main place errors creep in.

2. 2

   Distribute $-2x^2$. $-2x^2\cdot 9x^2=-18x^4$, $-2x^2\cdot(-4x)=+8x^3$, and $-2x^2\cdot 2=-4x^2$.

3. 3

   Distribute $+7$. $7\cdot 9x^2=63x^2$, $7\cdot(-4x)=-28x$, and $7\cdot 2=14$.

4. 4

   Line up all six partial products. $-18x^4+8x^3-4x^2+63x^2-28x+14$.

5. 5

   Combine the only pair of like terms. The two $x^2$ terms merge: $-4x^2+63x^2=59x^2$. Every other degree appears once, so the result is $-18x^4+8x^3+59x^2-28x+14$.

6. 6

   Verify numerically. At $x=1$: $(-2+7)(9-4+2)=5\cdot 7=35$, and $-18+8+59-28+14=35$. At $x=-1$: $(-2+7)(9+4+2)=5\cdot 15=75$, and $-18-8+59+28+14=75$.