\dfrac{3n^4-12n^2x^2}{8nx^2-4n^2x} \cdot \dfrac{-8x}{n^2-2nx-8x^2}=

asked by guest
on Jan 27, 2025 at 3:22 am



You asked:

Evaluate the expression: \(\frac{3 {n}^{4} - 12 {n}^{2} {x}^{2}}{8 n {x}^{2} - 4 {n}^{2} x} \cdot \frac{-\left( 8 x \right)}{{n}^{2} - 2 n x - 8 {x}^{2}}\)

MathBot Answer:

Evaluated



\(\displaystyle \frac{3 {n}^{4} - 12 {n}^{2} {x}^{2}}{8 n {x}^{2} - 4 {n}^{2} x} \cdot \frac{-\left( 8 x \right)}{{n}^{2} - 2 n x - 8 {x}^{2}} = - \frac{8 x \left(3 n^{4} - 12 n^{2} x^{2}\right)}{\left(- 4 n^{2} x + 8 x^{2} n\right) \left(n^{2} - 2 n x - 8 x^{2}\right)} \)

Expanded

\[\frac{3 {n}^{4} - 12 {n}^{2} {x}^{2}}{8 n {x}^{2} - 4 {n}^{2} x} \cdot \frac{-\left( 8 x \right)}{{n}^{2} - 2 n x - 8 {x}^{2}} = - \frac{24 n^{4} x}{- 4 n^{4} x + 16 n^{3} x^{2} + 16 n^{2} x^{3} - 64 x^{4} n} + \frac{96 n^{2} x^{3}}{- 4 n^{4} x + 16 n^{3} x^{2} + 16 n^{2} x^{3} - 64 x^{4} n}\]

Factored

\[\frac{3 {n}^{4} - 12 {n}^{2} {x}^{2}}{8 n {x}^{2} - 4 {n}^{2} x} \cdot \frac{-\left( 8 x \right)}{{n}^{2} - 2 n x - 8 {x}^{2}} = \frac{6 n}{n - 4 x}\]

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