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February 27th, 2011, 03:00 AM   #1
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Factorise cubic equation

Question 1
Solve the following cubic equations:
6x-17x-26x+8 = 0

My answer:
Step 1:Let f(x) =6x-17x-26x+8
When x=, f() = 6()+17()-26()+8
= 0
Step 2: From factor theorem, (2x-1) is a factor of f(x).
Using long division method,
X=, x=?, x=-4
Question 2
Solve the following cubic equations:
20x-68x+69x-18 = 0
My problem:
Im having problem in doing step one in which I have to find suitable value to replace x so that the equation becomes 0. Sure there are other methods to find the value of x in step one rather than trying to substitute the x with many values until we get the correct value. Sure there must be more systematic way than the approach in step one right? Can anyone guide me on that? Appreciate on your helps.
Merain is offline  
February 27th, 2011, 03:32 AM   #2
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Re: Factorise cubic equation

If a rational number is a root of the polynomial, then p is a factor of the constant term (in this case -1 and q is a factor of the leading term (in this case (20). So the possible positive values for p are 1, 2, 3, 6, 9, 18. The possible values for q are 1, 2, 4, 5, 10, 20. So just start doing synthetic division with the values for and their negatives until you get 0. You can do this very quickly using a single table if you do the multiplication and addition in your head before writing down each number. If none of these rational numbers are solutions, then all the roots are irrational or imaginary and you can probably only approximate them.
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March 8th, 2011, 10:37 PM   #3
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Re: Factorise cubic equation

Thnx DrSteve... it's much easier and fast using your method. Sorry for delay in my reply coz I was busy with my studies.
Merain is offline  
March 11th, 2011, 07:15 PM   #4
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Can you determine any integer for x that is less than or equal to 8 that is divisible by 8 such that 6x - 17x - 26x + 8 = 0? If so, use that integer such that (quadratic expression)(kx - integer) = 6x - 17x - 26x + 8 = 0, where k is real.
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