Multiply out (8y^4 - 3y^2 + 3y^3 - 8x^2y - 5y + 3 - 8x) times 7y^2.

Step-by-step solution

1. 1

   Set up the distribution and count the terms. The distributive law says every one of the seven terms inside the bracket gets multiplied by $7y^2$, so the answer will have seven terms. Counting first is a cheap guard against dropping one — the usual mistake with a bracket this long.

2. 2

   Recall the two rules for each product. Multiply the numerical coefficients, and add the exponents on any shared variable: $y^m\cdot y^2=y^{m+2}$. A term with no $y$ at all, like $3$ or $-8x$, simply gains a factor $y^2$. Signs travel with their term.

3. 3

   Multiply the pure powers of $y$.

   $8y^4\cdot 7y^2=56y^{6},\qquad 3y^3\cdot 7y^2=21y^{5},\qquad -3y^2\cdot 7y^2=-21y^{4},\qquad -5y\cdot 7y^2=-35y^{3}$

4. 4

   Multiply the terms that also contain $x$, plus the constant. The $x$ factors are untouched because $7y^2$ has no $x$:

   $-8x^2y\cdot 7y^2=-56x^2y^{3},\qquad -8x\cdot 7y^2=-56xy^{2},\qquad 3\cdot 7y^2=21y^{2}$

5. 5

   Collect the seven products in descending powers of $y$.

   $56y^6+21y^5-21y^4-56x^2y^3-35y^3+21y^2-56xy^2$

   Note that $-56x^2y^3$ and $-35y^3$ are not like terms (one carries $x^2$), and neither are $21y^2$ and $-56xy^2$, so nothing combines.

6. 6

   Check with a numeric substitution. At $x=1,y=1$ the bracket is $8-3+3-8-5+3-8=-10$, so the product should be $-10\cdot 7=-70$. Adding the answer terms: $56+21-21-56-35+21-56=-70$. ✓