Is it possible to factor $y = x^{3} + 4 x^{2} - x$ $y=x^3 + 4x^2 - x$ ? If so, what are the factors?

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Is it possible to factor $y = x^{3} + 4 x^{2} - x$ $y=x^3 + 4x^2 - x$ ? If so, what are the factors?

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Answer 1 · 2015-12-20T11:58:30

Yes, with irrational coefficients:

$x^{3} + 4 x^{2} - x = x (x + 2 - \sqrt{5}) (x + 2 + \sqrt{5})$ $x^3+4x^2-x = x(x+2-sqrt(5))(x+2+sqrt(5))$

Explanation:

Separate out the common factor $x$ $x$ of the terms, complete the square then use the difference of squares identity to finish.

The difference of squares identity can be written:

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

We use this below with $a = (x + 2)$ $a=(x+2)$ and $b = \sqrt{5}$ $b=sqrt(5)$

So we find:

$x^{3} + 4 x^{2} - x$ $x^3+4x^2-x$

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

$= x (x^{2} + 4 x + 4 - 5)$ $=x(x^2+4x+4-5)$

$= x ({(x + 2)}^{2} - {(\sqrt{5})}^{2})$ $=x((x+2)^2-(sqrt(5))^2)$

$= x ((x + 2) - \sqrt{5}) ((x + 2) + \sqrt{5})$ $=x((x+2)-sqrt(5))((x+2)+sqrt(5))$

$= x (x + 2 - \sqrt{5}) (x + 2 + \sqrt{5})$ $=x(x+2-sqrt(5))(x+2+sqrt(5))$

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Is it possible to factor $y = x^{3} + 4 x^{2} - x$ $y=x^3 + 4x^2 - x$ ? If so, what are the factors?

1 Answer

Explanation:

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