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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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High School Teacher
`(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`
`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)` :
`therefore` x=0 and x=6
Posted by durbanville on April 4, 2013 at 6:00 AM (Answer #1)
factor out x
either x=0 or `(x-6)^2=0`
Posted by pramodpandey on April 4, 2013 at 6:32 AM (Answer #2)
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