How do you factor $y = x^{3} + 3 x^{2} + 9 x + 27$ $y= x^3 + 3x^2 + 9x + 27$ ?

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How do you factor $y = x^{3} + 3 x^{2} + 9 x + 27$ $y= x^3 + 3x^2 + 9x + 27$ ?

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Answer 1 · 2018-05-05T07:03:57

$y = (x^{2} + 9) (x + 3)$ $y = (x^2+9)(x+3)$

$y = (x - 3 i) (x + 3 i) (x + 3)$ $color(white)(y) = (x-3i)(x+3i)(x+3)$

Explanation:

Given:

$y = x^{3} + 3 x^{2} + 9 x + 27$ $y = x^3+3x^2+9x+27$

Note that the ratio of the first and second terms is the same as that of the third and fourth terms.

So this cubic will factor by grouping:

$y = x^{3} + 3 x^{2} + 9 x + 27$ $y = x^3+3x^2+9x+27$

$y = (x^{3} + 3 x^{2}) + (9 x + 27)$ $color(white)(y) = (x^3+3x^2)+(9x+27)$

$y = x^{2} (x + 3) + 9 (x + 3)$ $color(white)(y) = x^2(x+3)+9(x+3)$

$y = (x^{2} + 9) (x + 3)$ $color(white)(y) = (x^2+9)(x+3)$

The remaining quadratic $x^{2} + 9$ $x^2+9$ is positive for all real values of $x$ $x$ , so has no linear factors with real coefficients.

It can be factored with complex coefficients as a difference of squares...

$x^{2} + 9 = x^{2} + 3^{2} = x^{2} - {(3 i)}^{2} = (x - 3 i) (x + 3 i)$ $x^2+9 = x^2+3^2 = x^2-(3i)^2 = (x-3i)(x+3i)$

where $i$ $i$ is the imaginary unit, satisfying $i^{2} = - 1$ $i^2=-1$

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How do you factor $y = x^{3} + 3 x^{2} + 9 x + 27$ $y= x^3 + 3x^2 + 9x + 27$ ?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you factor y= x^3 + 3x^2 + 9x + 27y=x3+3x2+9x+27y= x^3 + 3x^2 + 9x + 27?

1 Answer

Explanation:

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How do you factor $y = x^{3} + 3 x^{2} + 9 x + 27$ $y= x^3 + 3x^2 + 9x + 27$ ?