Is it possible to factor $y = x^{4} - 13 x^{2} + 36$ $y=x^4 - 13x^2 + 36$ ? If so, what are the factors?

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Is it possible to factor $y = x^{4} - 13 x^{2} + 36$ $y=x^4 - 13x^2 + 36$ ? If so, what are the factors?

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Answer 1 · 2018-04-16T08:52:54

$y = (x + 2) (x - 2) (x + 3) (x - 3)$ $y=(x+2)(x-2)(x+3)(x-3)$

Explanation:

$x^{4} - 13 x + 36$ $x^4-13x+36$

$u = x^{2}$ $u=x^2$

$u^{2} - 13 u + 36$ $u^2-13u+36$

a quadratic in u

$= (u - 4) (u - 9)$ $=(u-4)(u-9)$

$(x^{2} - 4) (x^{2} - 9)$ $(x^2-4)(x^2-9)$

both brackets are differences of squares

$(x + 2) (x - 2) (x + 3) (x - 3)$ $(x+2)(x-2)(x+3)(x-3)$

Answer 2 · 2018-04-16T09:03:40

$y = (x - 3) (x + 3) (x - 2) (x + 2)$ $y=(x-3)(x+3)(x-2)(x+2)$

Explanation:

$y = x^{4} - 13 x^{2} + 36$ $y=x^4-13x^2+36$ => Let: $x^{2} = u$ $x^2 = u$ , then: $x^{4} = u^{2}$ $x^4=u^2$ :
$y = u^{2} - 13 u + 36$ $y=u^2-13u+36$ => find 2 no's that multiply to 36, add to -13:
$y = u^{2} - 9 u - 4 u + 36$ $y=u^2-9u-4u+36$
$y = u (u - 9) - 4 (u - 9)$ $y=u(u-9)-4(u-9)$
$y = (u - 9) (u - 4)$ $y=(u-9)(u-4)$ => substitute back $x^{2} = u$ $x^2=u$ :
$y = (x^{2} - 9) (x^{2} - 4)$ $y=(x^2-9)(x^2-4)$ => factor using the difference of squares:
$y = (x - 3) (x + 3) (x - 2) (x + 2)$ $y=(x-3)(x+3)(x-2)(x+2)$

Answer 3 · 2018-04-16T09:05:15

$(x + 3) (x - 3) (x - 2) (x + 2)$ $(x+3)(x-3)(x-2)(x+2)$

Explanation:

$using the substitution u = x^{2}$ $"using the "color(blue)"substitution "u=x^2$

$\Rightarrow x^{4} - 13 x^{2} + 36 = u^{2} - 13 u + 36$ $rArrx^4-13x^2+36=u^2-13u+36$

$the factors of + 36 which sum to - 13 are - 9 and - 4$ $"the factors of + 36 which sum to - 13 are - 9 and - 4"$

$\Rightarrow u^{2} - 13 u + 36 = (u - 9) (u - 4)$ $rArru^2-13u+36=(u-9)(u-4)$

$change u back into terms of x gives$ $"change u back into terms of x gives"$

$(x^{2} - 9) (x^{2} - 4)$ $(x^2-9)(x^2-4)$

$both x^{2} - 9 and x^{2} - 4 are difference of squares$ $"both "x^2-9" and "x^2-4" are "color(blue)"difference of squares"$

$which factorise, in general as$ $"which factorise, in general as"$

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

$\Rightarrow x^{2} - 9 = (x - 3) (x + 3)$ $rArrx^2-9=(x-3)(x+3)$

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

$\Rightarrow x^{4} - 13 x^{2} + 36 = (x - 3) (x + 3) (x - 2) (x + 2)$ $rArrx^4-13x^2+36=(x-3)(x+3)(x-2)(x+2)$

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Is it possible to factor $y = x^{4} - 13 x^{2} + 36$ $y=x^4 - 13x^2 + 36$ ? If so, what are the factors?

3 Answers

Explanation:

Explanation:

Explanation:

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Is it possible to factor y=x4−13x2+36y=x^4 - 13x^2 + 36? If so, what are the factors?

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Explanation:

Explanation:

Explanation:

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Is it possible to factor $y = x^{4} - 13 x^{2} + 36$ $y=x^4 - 13x^2 + 36$ ? If so, what are the factors?