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September 3, 2014

September 3, 2014

Posted by **sarah** on Monday, March 2, 2009 at 6:03pm.

I used synthetic division and got 20x^2-6x-2=0

took 2 and divided it out of that equation

2(x+5/2)=10x^2 -3x-1

then i factored it out the right side of the equation

(x+5/2)=(5x+1)(2x-1)

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 6:24pmsince 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)(2x-1)

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)(2x-1) = 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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