2x^3-7x^2+x+6 one of the root is x=3 find the remaining roots. i got to this step 2x^2-x-2=0 i cant factor it so what do i do?

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2x^3-7x^2+x+6 one of the root is x=3 find the remaining roots. i got to this step 2x^2-x-2=0 i cant factor it so what do i do?

Mathematics
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if you know a root is 3 syndivide it 3 | 2 -7 1 6 0 6 -3 -6 ----------- 2 -1 -2 0 (2x^2 -x -2) = 0 2*2 = 4 1,4; 2,2 neither subtract to -1 so complete the square/ quad formula it
-mid# sqrt(mid#^2 - 4(first)(last)) ------ +- ----------------------- 2(first#) 2(first#)
1/4 +- sqrt(1+16)/4 1/4 +- sqrt(17)/4

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2x^2 -x -2 = 0 ------------- 2 x^2 -(1/2)x -1 = 0 x^2 -(1/2)x + ___ = 1 + ___ x^2 -1/2x + 1/16 = 1 + 1/16 (x-1/4)^2 = 17/16 x- 1/4 = +- sqrt(17/16) = +-sqrt(17)/4 x = 1/4 +- sqrt(17)/4
i mean what happen to -1 and how is 1/16 come from
a complete square can change back and forth like this: (x+2)^2 <--> (x^2 +4x +4) right?
an incomplete square is stuck... (x^2 +4x + ___ ) is stuck and cant change without first 'completing the square' right?
what we need are 2 numbers that add to 4 and multiply them together to get the missing term; but those 2 numbers have to be the SAME number. 3+1 cant work to complete the square 5-1 cant work to complete the square we need to identical numbers in order to complete the square... n+n = 4 2n = 4 n = 4/2 = 2 2+2 = 4; the missing term is the 2*2; which equals 4 right?
x^2 - 6x + ___ ??? we need to complete this square n+n = -6 2n = -6 n = -6/2 = -3 since -3+-3 =-6; the missing term is -3.-3 = 9 x^2 -6x +9 is a complete square right?
yup! fantastic thank you very much!!

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