How to factor $\frac{1}{2} x^{2} - \frac{1}{8}$ $1/2x^2 -1/8$ so that factors of two or more terms contain no fractional coefficients?

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How to factor $\frac{1}{2} x^{2} - \frac{1}{8}$ $1/2x^2 -1/8$ so that factors of two or more terms contain no fractional coefficients?

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Answer 1 · 2018-02-08T10:36:07

$\frac{1}{8} (2 x - 1) (2 x + 1)$ $1/8(2x-1)(2x+1)$

Explanation:

$take out a common factor of \frac{1}{8}$ $"take out a "color(blue)"common factor of "1/8$

$\Rightarrow \frac{1}{8} (4 x^{2} - 1)$ $rArr1/8(4x^2-1)$

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

$∙ x a^{2} - b^{2} = (a - b) (a + b)$ $•color(white)(x)a^2-b^2=(a-b)(a+b)$

$\Rightarrow 4 x^{2} - 1 = (2 x - 1) (2 x + 1)$ $rArr4x^2-1=(2x-1)(2x+1)$

$\Rightarrow \frac{1}{2} x^{2} - \frac{1}{8} = \frac{1}{8} (2 x - 1) (2 x + 1)$ $rArr1/2x^2-1/8=1/8(2x-1)(2x+1)$

Answer 2 · 2018-02-08T12:40:44

$\frac{1}{8}$ $1/8$ $(2 x + 1) (2 x - 1)$ $(2x+1)(2x-1)$

Explanation:

Factor:

$\frac{1}{2} x^{2} - \frac{1}{8}$ $1/2x^2-1/8$

Simplify $\frac{1}{2} x^{2}$ $1/2x^2$ to $\frac{x^{2}}{2}$ $(x^2)/2$ .

$\frac{x^{2}}{2} - \frac{1}{8}$ $x^2/2-1/8$

The common denominator is $8$ $8$ . Multiply $\frac{x^{2}}{2}$ $x^2/2$ by $\frac{4}{4}$ $4/4$ to convert the denominator to $8$ $8$ .

$\frac{x^{2}}{2} \times \frac{4}{4} - \frac{1}{8}$ $x^2/2xx4/4-1/8$

Simplify.

$\frac{4 x^{2}}{8} - \frac{1}{8}$ $(4x^2)/8-1/8$

Combine the numerators into a common fraction.

$\frac{4 x^{2} - 1}{8}$ $(4x^2-1)/8$

Express the numerator as a difference of squares:

$(a^{2} - b^{2}) = (a + b) (a - b)$ $(a^2-b^2)=(a+b)(a-b)$

${(2 x)}^{2} - 1^{2} = (2 x + 1) (2 x - 1)$ $(2x)^2-1^2=(2x+1)(2x-1)$

Place the factorization over the denominator $8$ $8$ .

$\frac{(2 x + 1) (2 x - 1)}{8}$ $((2x+1)(2x-1))/8$

Alternatively,

$\frac{1}{8}$ $1/8$ $(2 x + 1) (2 x - 1)$ $(2x+1)(2x-1)$

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How to factor $\frac{1}{2} x^{2} - \frac{1}{8}$ $1/2x^2 -1/8$ so that factors of two or more terms contain no fractional coefficients?

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How to factor $\frac{1}{2} x^{2} - \frac{1}{8}$ $1/2x^2 -1/8$ so that factors of two or more terms contain no fractional coefficients?