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Steps for factoring and then solving please: `(x-4)^3 -4(3x-16)=0` (x-4)^3 -4(3x-16)=0...

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aego-aguirre | eNoter

Posted April 4, 2013 at 5:23 AM via web

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Steps for factoring and then solving please:

`(x-4)^3 -4(3x-16)=0`

(x-4)^3 -4(3x-16)=0 (same thing typed)

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durbanville | High School Teacher | (Level 1) Educator Emeritus

Posted April 4, 2013 at 6:00 AM (Answer #1)

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`(x-4)^3 - 4(3x-16)=0`

Expand so that like terms can be arranged together to solve. As we have `(x-4)^3`  which may be difficult to factorize, do it in stages by doing `(x-4)^2` first:

`(x-4)(x^2 - 8x +16) - 12x + 64 = 0`

Note the changing of the symbol when multiplying the`-4 times -16=64`  

Now multiply the other `(x-4)` in:

`(x^3 - 8x^2 +16x - 4x^2+32x -64 ) - 12x +64=0`

Put like terms together and add/subract as required:

`x^3 -12x^2 +36x = 0`

Factorize:

`x(x^2 -12x +36) = 0`

`therefore x=0 and (x^2 -12x +36) = 0`

Factorize using the factors of the 1st and 3rd terms ie `x times x`  and the only factors that will work to render a middle term of -12 ie `(6 times 6)` :

`(x-6)(x-6)=0`

`therefore x=6`

`therefore` x=0 and x=6

Sources:

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pramodpandey | College Teacher | Valedictorian

Posted April 4, 2013 at 6:32 AM (Answer #2)

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`(x-4)^3-4(3x-16)=0`

Open brackets

`x^3-4^3-3x4(x-4)-12x+64=0`

`x^3-64-12x^2+48x-12x+64=0`

`x^3-12x^2+36x=0`

factor out x

`x (x^2-12x+36)=0`

`x(x^2-6x-6x+36)=0`

`x{x(x-6)-6(x-6)}=0`

`x{(x-6)(x-6)}=0`

`x(x-6)^2=0`

 This means

either x=0   or `(x-6)^2=0`

`x-6=0`

`x=6`

So answer

x=0,6

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