What is (x^2+1/4)(x^2+1/5) ?

The solution to (x^2+1/4)(x^2+1/5) is ((4x^2+1)(5x^2+1))/(20)

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simplify (x^2+1/4)(x^2+1/5)

simplify

(x^{2} + \frac{1}{4}) (x^{2} + \frac{1}{5})

\frac{( 4 x ^{2} + 1 ) ( 5 x ^{2} + 1 )}{20}

Solution steps

(x^{2} + \frac{1}{4}) (x^{2} + \frac{1}{5})

Simplify

x^{2} + \frac{1}{4} : \frac{4 x ^{2} + 1}{4}

= \frac{4 x ^{2} + 1}{4} (x^{2} + \frac{1}{5})

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 4 x ^{2} + 1 ) ( x ^{2} + \frac{1}{5} )}{4}

Simplify

(4 x^{2} + 1) (x^{2} + \frac{1}{5}) : \frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 5 + 1 )}{5}

= \frac{\frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 5 + 1 )}{5}}{4}

Apply the fraction rule:

\frac{\frac{a}{b}}{c} = \frac{a}{b \cdot c}

= \frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 5 + 1 )}{5 \cdot 4}

Multiply the numbers:

5 \cdot 4 = 20

= \frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 5 + 1 )}{20}

= \frac{( 4 x ^{2} + 1 ) ( 5 x ^{2} + 1 )}{20}

(x - 2) (x - 3) (x - 1)

(3 x^{\frac{5}{2}} - 7 y^{\frac{7}{2}}) (3 x^{\frac{5}{2}} + 7 y^{\frac{7}{2}})

\frac{135}{( 15 + x ) ^{2}}

(1 - t, - (1 - t)^{2} - t^{2}) \cdot (1 - t, t)

- 2 sin (3 \cdot 0) + 1

What is (x^2+1/4)(x^2+1/5) ?
The solution to (x^2+1/4)(x^2+1/5) is ((4x^2+1)(5x^2+1))/(20)