Solve left(k^2+1right)left(16k^2-100right)-left(2k^2+25right)16k^2/25left(4{k^2+1right)}+frac{4{k}^2}{25}

$$\frac{ \left( { k }^{ 2 } +1 \right) \left( 16 { k }^{ 2 } -100 \right) - \left( 2 { k }^{ 2 } +25 \right) 16 { k }^{ 2 } }{ 25 \left( 4 { k }^{ 2 } +1 \right) } + \frac{ 4 { k }^{ 2 } }{ 25 }$$

Evaluate

$$-\frac{4\left(24k^{2}+5\right)}{5\left(4k^{2}+1\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 25\left(4k^{2}+1\right) and 25 is 25\left(4k^{2}+1\right). Multiply \frac{4k^{2}}{25} times \frac{4k^{2}+1}{4k^{2}+1}.

$$\frac{\left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}}{25\left(4k^{2}+1\right)}+\frac{4k^{2}\left(4k^{2}+1\right)}{25\left(4k^{2}+1\right)}$$

Since \frac{\left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}}{25\left(4k^{2}+1\right)} and \frac{4k^{2}\left(4k^{2}+1\right)}{25\left(4k^{2}+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}+4k^{2}\left(4k^{2}+1\right)}{25\left(4k^{2}+1\right)}$$

Do the multiplications in \left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}+4k^{2}\left(4k^{2}+1\right).

$$\frac{16k^{4}-100k^{2}+16k^{2}-100-\left(32k^{4}+400k^{2}\right)+16k^{4}+4k^{2}}{25\left(4k^{2}+1\right)}$$

Combine like terms in 16k^{4}-100k^{2}+16k^{2}-100-\left(32k^{4}+400k^{2}\right)+16k^{4}+4k^{2}.

$$\frac{-480k^{2}-100}{25\left(4k^{2}+1\right)}$$

Factor the expressions that are not already factored in \frac{-480k^{2}-100}{25\left(4k^{2}+1\right)}.

$$\frac{20\left(-24k^{2}-5\right)}{25\left(4k^{2}+1\right)}$$

Cancel out 5 in both numerator and denominator.

$$\frac{4\left(-24k^{2}-5\right)}{5\left(4k^{2}+1\right)}$$

Expand 5\left(4k^{2}+1\right).

$$\frac{4\left(-24k^{2}-5\right)}{20k^{2}+5}$$

Use the distributive property to multiply 4 by -24k^{2}-5.

$$\frac{-96k^{2}-20}{20k^{2}+5}$$

Expand

$$-\frac{4\left(24k^{2}+5\right)}{5\left(4k^{2}+1\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 25\left(4k^{2}+1\right) and 25 is 25\left(4k^{2}+1\right). Multiply \frac{4k^{2}}{25} times \frac{4k^{2}+1}{4k^{2}+1}.

$$\frac{\left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}}{25\left(4k^{2}+1\right)}+\frac{4k^{2}\left(4k^{2}+1\right)}{25\left(4k^{2}+1\right)}$$

Since \frac{\left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}}{25\left(4k^{2}+1\right)} and \frac{4k^{2}\left(4k^{2}+1\right)}{25\left(4k^{2}+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}+4k^{2}\left(4k^{2}+1\right)}{25\left(4k^{2}+1\right)}$$

Do the multiplications in \left(k^{2}+1\right)\left(16k^{2}-100\right)-\left(2k^{2}+25\right)\times 16k^{2}+4k^{2}\left(4k^{2}+1\right).

$$\frac{16k^{4}-100k^{2}+16k^{2}-100-\left(32k^{4}+400k^{2}\right)+16k^{4}+4k^{2}}{25\left(4k^{2}+1\right)}$$

Combine like terms in 16k^{4}-100k^{2}+16k^{2}-100-\left(32k^{4}+400k^{2}\right)+16k^{4}+4k^{2}.

$$\frac{-480k^{2}-100}{25\left(4k^{2}+1\right)}$$

Factor the expressions that are not already factored in \frac{-480k^{2}-100}{25\left(4k^{2}+1\right)}.

$$\frac{20\left(-24k^{2}-5\right)}{25\left(4k^{2}+1\right)}$$

Cancel out 5 in both numerator and denominator.

$$\frac{4\left(-24k^{2}-5\right)}{5\left(4k^{2}+1\right)}$$

Expand 5\left(4k^{2}+1\right).

$$\frac{4\left(-24k^{2}-5\right)}{20k^{2}+5}$$

Use the distributive property to multiply 4 by -24k^{2}-5.

$$\frac{-96k^{2}-20}{20k^{2}+5}$$