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Finding the Quotient (18x3y36xy2)÷(3xy)(18x^3y - 36xy^2) \div (-3xy)

Divide a two-variable binomial by a negative two-variable monomial: (18x3y36xy2)÷(3xy)(18x^3y - 36xy^2) \div (-3xy).

Step 1 — Rewrite as a fraction. Place the polynomial in the numerator and the monomial in the denominator: 18x3y36xy23xy\frac{18x^3y - 36xy^2}{-3xy}.

Step 2 — Separate the terms. Split the single fraction into two individual fractions by dividing each term of the numerator by the denominator: 18x3y3xy36xy23xy\frac{18x^3y}{-3xy} - \frac{36xy^2}{-3xy}.

Step 3 — Simplify each fraction. For the first term: divide the coefficients 183=6\frac{18}{-3} = -6, apply the Quotient Property for xx to get x3x=x31=x2\frac{x^3}{x} = x^{3-1} = x^2, and for yy to get yy=1\frac{y}{y} = 1. The simplified first term is 6x2-6x^2. For the second term: divide the coefficients 363=12\frac{36}{-3} = -12, for xx we have xx=1\frac{x}{x} = 1, and for yy we have y2y=y21=y\frac{y^2}{y} = y^{2-1} = y. The simplified second term is 12y-12y. Because the original expression subtracts this term, it becomes (12y)-(-12y), which simplifies to +12y+12y.

The quotient is 6x2+12y-6x^2 + 12y.

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Updated 2026-04-29

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