SOLUTION: Suppose that 1, 2, and 3 are roots of the degree five polynomial equation, x^5 -8x^4 + 15x^3 + 20x^2-76x + 48 = 0. The equation has two additional roots. What are they? (Hint: Av

Question 896827: Suppose that 1, 2, and 3 are roots of the degree five polynomial equation,
x^5 -8x^4 + 15x^3 + 20x^2-76x + 48 = 0.
The equation has two additional roots. What are they? (Hint: Avoid long division.)

Found 2 solutions by josgarithmetic, richwmiller:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
Your instructions suggest not using LONG DIVISION; but hopefully you are allowed to use SYNTHETIC DIVISION.

The possible roots to check for (using synthetic division) would be, other than the 1, 2, and 3; plus-and-minus of 4, 8, 12, 24 and maybe 48. You are probably more interested in the negative roots, since there are a couple of sign changes in that degree-5 polynomial. You might want to check -1, -2, or -3.

If you would perform the synthetic divisions for the given known roots, you would obtain a degree-two polynomial, easily analyzed for roots using general solution for quadratic equation.

I only described what could or should be done; not actually did it.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!

x^5 − 8x^4 + 15x^3 + 20x^2 − 76x + 48 = 0

48/6=8 so we can eliminate 12, 24 and 48 want to check factors of 8 that is 8,4,-1, -2
just plug them in and see if they work
(4)^5 − 8(4)^4 + 15(4)^3 + 20(4)^2 − 76(4) + 48 = 0 true
(-2)^5 − 8(-2)^4 + 15(-2)^3 + 20(-2)^2 − 76(-2) + 48 = 0 true
(8)^5 − 8(8)^4 + 15(8)^3 + 20(8)^2 − 76(8) + 48 = 0 false
other roots are x=-2 and x=4

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