Question 29167: given a polynomial and one of its factors, find the remaining factors of the polynomial. Some factors may not be binomials.
6x^3-25x^2+2x+8;3x-2
Answer by Nate(3500)   (Show Source):
You can put this solution on YOUR website! Your Question:
given a polynomial and one of its factors, find the remaining factors of the polynomial. Some factors may not be binomials.
6x^3-25x^2+2x+8;3x-2
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My answer:
What i am going to show you is called synthetic division....
You're going to set 3x-2 to 0 to get x equaling a number
3x-2=0 (add 2)
3x=2 (divide by 3)
x=2/3
Now, take the coefficients (the number in front of the variables) and set them in a side-by-side order with your x value (2/3) on the way left(if you had a 0x^2, you would still use the 0 coefficient.)
(2/3)__6__(-25)__2__8
6 (multiply the 2/3 and 6 and add to -25)
-21 (multiply the 2/3 and -21 and add to 2)
-12 (multiply the 2/3 and -12 and add to 8)
0
Notice that you have 6, then, -21, and last -12, so you would combine them with the degree of x is descending order and multiply by the value you were given beforehand.
To simplify it more:
divide by 3 to make it easier------ (6x^2-21x-12)(x-2/3)
3(2x^2-7x-4)(x-2/3)------ multiply by the coefficient of x^2 and the constant(the number without a variable(the number is -8)) Then, find the factors of that number (1,2,4,8,-1,-2,-4,-8). Next, find the factors that would add to equal the -7 in the equation (the answer would be -8 and 1)
plug in the -8 and 1 to equal the -7 ------- 3(2x^2-8x+x-4)(x-2/3)
simplify 3{2x(x-4)+1(x-4)}(x-2/3)
combine 3(2x+1)(x-4)(x-2/3)
multiply 3 by one of the polynomials (6x+3)(x-4)(x-2/3)
(6x+3)(x-4)(x-2/3) is your answer!
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