Posted by sarah on Monday, March 2, 2009 at 6:03pm.
solve the equation 20x^3 +44x^2 17x 5 given that 5/2 is a zero of f(x) 20x^3 +44x^2 17x 5
I used synthetic division and got 20x^26x2=0
took 2 and divided it out of that equation
2(x+5/2)=10x^2 3x1
then i factored it out the right side of the equation
(x+5/2)=(5x+1)(2x1)
then i solved for x on each and got
5/2, 1/5, 1/2
is this correct?

Math  Reiny, Monday, March 2, 2009 at 6:24pm
since x = 5/2 , one of the factors had to be (2x+5)
when you divide f(x) by 2x+5 you get
10x^2  3x  1, not 20x^2  6x  2
I know that when you set your answer of
20x^2  6x  2 = 0 you get the same as my result when I set it mine to zero.
You have to be careful when using synthetic division by dividing polynomials by factors of the form ax ± b
it is safer to use long division for that.
you then make the statement
(x+5/2)=(5x+1)(2x1)
that is just wrong, they are not "equal"
what you have to say is
20x^3 +44x^2 17x 5 = 0
(2x+5)(10x^2  3x  1) = 0
(2x+5)(5x+1)(2x1) = 0
then x = 5/2 , we knew that, or
x = 1/5 or x = 1/2
I know you had those answers, but with me you would lose marks for bad form in the solution.
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