Question

How to factor `1/2x^2 -1/8` so that factors of two or more terms contain no fractional coefficients?

Answer 1

`1/8(2x-1)(2x+1)`

Explanation:

`"take out a "color(blue)"common factor of "1/8`

`rArr1/8(4x^2-1)`

`4x^2-1" is a "color(blue)"difference of squares"`

`•color(white)(x)a^2-b^2=(a-b)(a+b)`

`rArr4x^2-1=(2x-1)(2x+1)`

`rArr1/2x^2-1/8=1/8(2x-1)(2x+1)`

Answer 2

`1/8` `(2x+1)(2x-1)`

Explanation:

Factor:

`1/2x^2-1/8`

Simplify `1/2x^2` to `(x^2)/2`.

`x^2/2-1/8`

The common denominator is `8`. Multiply `x^2/2` by `4/4` to convert the denominator to `8`.

`x^2/2xx4/4-1/8`

Simplify.

`(4x^2)/8-1/8`

Combine the numerators into a common fraction.

`(4x^2-1)/8`

Express the numerator as a difference of squares:

`(a^2-b^2)=(a+b)(a-b)`

`(2x)^2-1^2=(2x+1)(2x-1)`

Place the factorization over the denominator `8`.

`((2x+1)(2x-1))/8`

Alternatively,

`1/8` `(2x+1)(2x-1)`