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Divide. If there is a remainder, include it as a simplified fraction. $\newline$ $(-30y^3 + 35y^2) \div 5y^2$

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Divide polynomials by monomials

Full solution

Q. Divide. If there is a remainder, include it as a simplified fraction.

\newline

(-30y^3 + 35y^2) \div 5y^2

Divide by Monomial: We have: $\newline$ $(-30y^3 + 35y^2) \div 5y^2$ $\newline$ First, we will divide each term in the polynomial by the monomial. $\newline$ $(-30y^3 + 35y^2) \div 5y^2$ $\newline$ $= \frac{-30y^3}{5y^2} + \frac{35y^2}{5y^2}$
Divide $-30y^3$ by $5y^2$ : What is $-30y^3$ divided by $5y^2$ ? $\newline$ (-30y^3)/(5y^2) \(\newline= -30/5 \times y^3/y^2 $\newline$ = -6 \times y^{(3-2)} $\newline$ = -6y\)
Divide $35y^2$ by $5y^2$ : What is $35y^2$ divided by $5y^2$ ? $\newline$ $(35y^2)/(5y^2)$ $\newline$ $= 35/5 \times y^2/y^2$ $\newline$ $= 7 \times 1$ $\newline$ $= 7$
Combine Results: We have: $\newline$ $(-30y^3)/(5y^2) = -6y$ $\newline$ $(35y^2)/(5y^2) = 7$ $\newline$ Now, write the result of $(-30y^3)/(5y^2) + (35y^2)/(5y^2)$ . $\newline$ $(-30y^3 + 35y^2) \div 5y^2$ $\newline$ $= (-30y^3)/(5y^2) + (35y^2)/(5y^2)$ $\newline$ $= -6y + 7$

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