Question

$\frac{28m^{2}n^{2}+35m^{3}n - 36mn^{2}}{14m^{2}n^{2}}=square$ (use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Factor out the greatest - common factor in the numerator

The GCF of \(28m^{2}n^{2}\), \(35m^{3}n\) and \(- 35mn^{2}\) is \(7mn\). So, \(28m^{2}n^{2}+35m^{3}n - 35mn^{2}=7mn(4mn + 5m^{2}-5n)\).

Step2: Rewrite the original expression

The original expression \(\frac{28m^{2}n^{2}+35m^{3}n - 35mn^{2}}{14m^{2}n^{2}}\) becomes \(\frac{7mn(4mn + 5m^{2}-5n)}{14m^{2}n^{2}}\).

Step3: Simplify the fraction

Cancel out the common factors. We can cancel out a factor of \(7mn\) from the numerator and denominator. \(\frac{7mn(4mn + 5m^{2}-5n)}{14m^{2}n^{2}}=\frac{4mn + 5m^{2}-5n}{2mn}\).

Step4: Split the fraction into three terms

\(\frac{4mn + 5m^{2}-5n}{2mn}=\frac{4mn}{2mn}+\frac{5m^{2}}{2mn}-\frac{5n}{2mn}\).

Step5: Simplify each term

\(\frac{4mn}{2mn}=2\), \(\frac{5m^{2}}{2mn}=\frac{5m}{2n}\), \(\frac{5n}{2mn}=\frac{5}{2m}\). So the result is \(2+\frac{5m}{2n}-\frac{5}{2m}\).

Answer:

\(2+\frac{5m}{2n}-\frac{5}{2m}\)