What is (x^2+1/4)(x^2-3/4) ?

The solution to (x^2+1/4)(x^2-3/4) is ((4x^2+1)(4x^2-3))/(16)

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simplify (x^2+1/4)(x^2-3/4)

simplify

(x^{2} + \frac{1}{4}) (x^{2} - \frac{3}{4})

\frac{( 4 x ^{2} + 1 ) ( 4 x ^{2} - 3 )}{16}

Solution steps

(x^{2} + \frac{1}{4}) (x^{2} - \frac{3}{4})

Simplify

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

= \frac{4 x ^{2} + 1}{4} (x^{2} - \frac{3}{4})

Apply the fraction rule:

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

= \frac{( 4 x ^{2} + 1 ) ( x ^{2} - \frac{3}{4} )}{4}

Simplify

(4 x^{2} + 1) (x^{2} - \frac{3}{4}) : \frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 4 - 3 )}{4}

= \frac{\frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 4 - 3 )}{4}}{4}

Apply the fraction rule:

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

= \frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 4 - 3 )}{4 \cdot 4}

Multiply the numbers:

4 \cdot 4 = 16

= \frac{( 4 x ^{2} + 1 ) ( x ^{2} \cdot 4 - 3 )}{16}

= \frac{( 4 x ^{2} + 1 ) ( 4 x ^{2} - 3 )}{16}

(2 cos (A) + 1) (2 cos (A) - 1) (2 cos (2 A) - 1)

(t - 3) cos (t - 3)

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

(3 + \frac{i}{n})^{2}

tan (θ) cos (θ)

What is (x^2+1/4)(x^2-3/4) ?
The solution to (x^2+1/4)(x^2-3/4) is ((4x^2+1)(4x^2-3))/(16)