Simplify 7 x 11 x x x y x x y x squared.

Step-by-step solution

1. 1

   Reorder the factors by kind. Multiplication is commutative and associative, so the scattered $x$ and $y$ factors can be gathered together: $(7\cdot11)\cdot\left(x\cdot x\cdot x\cdot x^2\right)\cdot\left(y\cdot y\right).$ Grouping first is what prevents miscounting the exponents.

2. 2

   Multiply the numerical coefficients. $7 \cdot 11 = 77.$

3. 3

   Add the exponents of x. There are three bare $x$ factors (each $x^1$) plus one $x^2$, so by the product rule $a^m a^n = a^{m+n}$: $x^1\cdot x^1\cdot x^1\cdot x^2 = x^{1+1+1+2} = x^5.$ Counting the bare $x$'s carefully is the whole difficulty - they appear in positions 3, 4 and 6 of the original chain.

4. 4

   Add the exponents of y. Two bare $y$ factors give $y^{1+1} = y^2$.

5. 5

   Write the single monomial. $7 \cdot 11 \cdot x \cdot x \cdot y \cdot x \cdot y \cdot x^2 = 77x^5y^2.$ Standard form lists the coefficient first, then the variables alphabetically.

6. 6

   Verify with a numerical substitution. At $x=2$, $y=3$: the original is $7\cdot11\cdot2\cdot2\cdot3\cdot2\cdot3\cdot4 = 77\cdot288 = 22176$, and $77\cdot2^5\cdot3^2 = 77\cdot32\cdot9 = 22176$. They match, so the exponent count is right.