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Math - polynomials

  • Thread starter denian
  • Start date
  • #1
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given that f(x) = x^4 - 27x^2 - 14x + 120 can be expressed as

( x^2 + a )^2 - ( bx + 7 )^2

where a,b are constant. find the values of a and b. hence, or otherwise, factorise f(x) completely.




the value of a and b are -13 and 1 respectively.
so,

f(x) = ( x^2 - 13 )^2 - ( x + 7 )^2

one way to factorise f(x) completely is to substitute value of x so that we can get
f(x) = 0.

is there any other easier way since there is a keyword "HENCE" in the question.
thank you.
 

Answers and Replies

  • #2
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f(x) = 0
<=>
( x^2 - 13 )^2 = ( x + 7 )^2
Taking square roots on both sides:
|x^2 - 13| = |x + 7|
This breaks up into 2 separate equations which are just quadratic and, thus, can directly be solved, giving 2 zeroes of f each.
 
  • #3
HallsofIvy
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I suspect what was intended here was that you use

a2- b2= (a-b)(a+b).

Once you know that
x4 - 27x2 - 14x + 120= ( x2 - 13 )2 - ( x + 7 )2,

you can continue as
(x2- 13+ (x- 7))(x2-13-(x-7))
= (x2+ x- 20)(x2-x- 6)

Now can you further factor those two factors?
 
  • #4
HallsofIvy
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"That's Quaint"? :)
 

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