Find the product (3x^2 + 4xy + 5y^2)(7x^2 + 9y^2), writing the multiplication horizontally.

Step-by-step solution

1. 1

   Notice the structure. Both factors are homogeneous — every term of the first has total degree $2$ and every term of the second has total degree $2$ — so every partial product has total degree $4$. That makes $x^4,\ x^3y,\ x^2y^2,\ xy^3,\ y^4$ the only possible slots and gives a useful self-check.

2. 2

   Distribute the trinomial's terms across $7x^2$. $3x^2\cdot 7x^2=21x^4$, $4xy\cdot 7x^2=28x^3y$, and $5y^2\cdot 7x^2=35x^2y^2$.

3. 3

   Distribute across $9y^2$. $3x^2\cdot 9y^2=27x^2y^2$, $4xy\cdot 9y^2=36xy^3$, and $5y^2\cdot 9y^2=45y^4$.

4. 4

   Group the six monomials by pattern. $21x^4$ and $28x^3y$ and $36xy^3$ and $45y^4$ each stand alone; the two $x^2y^2$ terms are $35x^2y^2$ and $27x^2y^2$.

5. 5

   Combine and write the answer in descending powers of $x$. $35x^2y^2+27x^2y^2=62x^2y^2$, giving $21x^4+28x^3y+62x^2y^2+36xy^3+45y^4$.

6. 6

   Verify at two points. At $x=y=1$: $(3+4+5)(7+9)=12\cdot 16=192$ and $21+28+62+36+45=192$. At $x=2,y=-1$: $(12-8+5)(28+9)=9\cdot 37=333$ and $336-224+248-72+45=333$.