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How do you factor this equation?: `(x^4 - 3x^3 - 7x - 14)/(x-2)` 

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rosey-girl | Student, Grade 12 | Valedictorian

Posted May 8, 2013 at 11:17 PM via web

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How do you factor this equation?:

`(x^4 - 3x^3 - 7x - 14)/(x-2)` 

Tagged with algebra 2, math

2 Answers | Add Yours

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oldnick | Valedictorian

Posted May 9, 2013 at 12:18 AM (Answer #1)

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`(x^4-3x^3-7x-14)/(x-2)` `=(P(x))/(x-2)`

Since  `P(2)!=0`  `P(x)`  and `x-2`  have no common factors.

`P(x)= x^4-8x^3+24x^2-32x+16 +(5x^3-24x^2+25x-30)`

`=(x-2)^4+5x^3-24x^2+25x-30=`

`=(x-2)^4+ 5(x^3-5x^2+5x-6)+x^2=`

`=(x-2)^4 +5(x^3-6x^2+12x-8 +x^2-7x+2)+x^2=`

`=(x-2)^4+5(x-2)^3+5(x^2-7x+2) +x^2=`

`=(x-2)^4 +5(x-2)^3 + 6x^2-35x+10=`

`=(x-2)^4+5(x-2)^3+6(x^2-6x+4)+x-14=`

`=(x-2)^4+5(x-2)^3 +6(x^2-4x+4-2x)+x-14=`

`=(x-2)^4+5(x-2)^3+6[(x-2)^2-2x]+x-14=`

`=(x-2)^4+5(x-2)^3+6(x-2)^2-11x-14=`

`=(x-2)^4+5(x-2)^3+6(x-2)^2-11(x-2)-36`

So that:  `(x^4-3x^3-7x-14)/(x-2)=`

`=(x-2)^3+5(x-2)^2+6(x-2)-11-36/(x-2)`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted May 9, 2013 at 5:15 AM (Answer #2)

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You may use reminder theorem such that:

`x^4 - 3x^3 - 7x - 14 = (x - 2)*q(x) + r(x)`

`q(x)` represent the quotient and it is a polynomial whose order is 3

`r(x)` represents the reminder and its order needs to be smaller than the order of binomial `x - 2` , hence `r(x)` is a constant

Supposing that `q(x) = ax^3 + bx^2+ cx + d ` and `r(x) = e,` yields:

`x^4 - 3x^3 - 7x - 14 = (x - 2)*(ax^3 + bx^2+ cx + d) + e`

`x^4 - 3x^3 - 7x - 14 = ax^4 + bx^3 + cx^2 + dx - 2ax^3 - 2bx^2 - 2cx - 2d + e`

`x^4 - 3x^3 - 7x - 14 = ax^4 + x^3(b - 2a) + x^2(c - 2b) + x(d - 2c) - 2d + e`

Equating the coefficients of like powers yields:

`{(a = 1),(b - 2a = -3 => b = 1),(c - 2b = 0 => c = 2),(d - 2c = -7 => d = -3),(-2d + e = -14 => e = e = -20):}`

Replacing back the values for coefficients a,b,c,d,e, yields:

`x^4 - 3x^3 - 7x - 14 = (x - 2)*(x^3 + x^2+ 2x - 3) - 20`

Forming the original fraction yields:

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x^3 + x^2+ 2x - 3) - 20/(x - 2)`

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x^3 + x^2+ 2x - 2 - 1) - 20/(x - 2)`

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x^3 - 1) + (x^2 - 1) + (2x - 2) + 1 - 20/(x - 2)`

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x - 1)(x^2 + x + 1) + (x - 1)(x + 1) + 2(x - 1) + 1 - 20/(x - 2)`

Factoring out `x - 1` yields:

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x - 1)(x^2 + x + 1 + x + 1 + 2) + 1 - 20/(x - 2)`

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x - 1)(x^2 + 2x + 4) + 1 - 20/(x - 2)`

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x - 1)(x^2 + 2x + 4) + 1 - 20/(x - 2)`

`(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x - 1)(x^2 + 2x + 4) + (x - 22)/(x - 2)`

Hence, evaluating one factored form, using reminder theorem, yields `(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x - 1)(x^2 + 2x + 4) + (x - 22)/(x - 2).`

(x^4 - 3x^3 - 7x - 14)/(x - 2) = (x - 1)(x^2 + 2x + 4) + 1 - 20/(x - 2)

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