How do you factor $g (1 - x) - g x + g x^{2}$ $g(1-x) - gx + gx^2$ ?

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Answer 1 · 2018-03-30T01:02:54

We have:

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

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

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

Hopefully this helps!

Answer 2 · 2018-03-30T01:04:12

${g (x - 1)}^{2}$ $g(x-1)^2$

$g (1 - x) - g x + g x^{2}$ $g(1-x)-gx+gx^2$

Distribute:
$g - g x - g x + g x^{2}$ $g-gx-gx+gx^2$

Combine like terms:
$g - 2 g x + g x^{2}$ $g-2gx+gx^2$

Factor out $g$ $g$ with GCF:
$g (1 - 2 x + x^{2})$ $g(1-2x+x^2)$

Factor the quadratic:
$g (x - 1) (x - 1)$ $g(x-1)(x-1)$

Express in simplest form:
${g (x - 1)}^{2}$ $g(x-1)^2$

How do you factor g(1-x) - gx + gx^2g(1−x)−gx+gx2g(1-x) - gx + gx^2?