SOLUTION: P(x) = x^4 + 2x^3 - 2x^2 - 3x +2 Solve for p(x) =0

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Question 1180384: P(x) = x^4 + 2x^3 - 2x^2 - 3x +2
Solve for p(x) =0
Found 2 solutions by ewatrrr, MathLover1:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
1 and -2 are roots:
Using Synthetic Division:

1	1	2	-2	-3	2
		1	3	1	-2
-2	1	3	1	-2	0
		-2	-2	2	
	1	1	-1	0	
        x^2  +  x  -1 = 0      x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
 
   x+=+%28-1+%2B-+sqrt%285%29%29%2F%282%29+
      x = .618  x = -1.618  (rounded to 4 decimal places)
 x = { -2, -1.618, .618, 1}
Wish You the Best in your Studies.



Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

P%28x%29+=+x%5E4+%2B+2x%5E3+-+2x%5E2+-+3x+%2B2

x%5E4+%2B+2x%5E3+-+2x%5E2+-+3x+%2B2=0...factor
x%5E4+-x%5E3%2B3x%5E3-+3x%5E2%2Bx%5E2-x-+2x%2B2=0
%28x%5E4+-x%5E3%29%2B%283x%5E3-+3x%5E2%29%2B%28x%5E2-x%29-+%282x-2%29=0
x%5E3%28x+-1%29%2B3x%5E2%28x-+1%29%2Bx%28x-1%29-+2%28x-1%29=0
%28x+-+1%29+%28x%5E3+%2B+3x%5E2+%2B+x+-+2+%29=+0
%28x+-+1%29+%28x%5E3++%2B2x%5E2%2B+x%5E2%2B+2x-x++-+2+%29=+0
%28x+-+1%29+%28%28x%5E3++%2B2x%5E2%29%2B+%28x%5E2%2B+2x%29-%28x+%2B+2%29+%29=+0
%28x+-+1%29+%28x%5E2%28x++%2B2%29%2Bx+%28x%2B+2%29-%28x+%2B+2%29+%29=+0
%28x+-+1%29+%28x+%2B+2%29+%28x%5E2+%2B+x+-+1%29+=+0
solutions:
%28x+-+1%29++=+0=>x=1
%28x+%2B+2%29=+0=>x=-2
+%28x%5E2+%2B+x+-+1%29+=+0...use quadratic formula
x=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29
x=%28-1%2B-sqrt%281%5E2-4%2A1%2A%28-1%29%29%29%2F%282%2A1%29
x=%28-1%2B-sqrt%281%2B4%29%29%2F2
x=%28-1%2B-sqrt%285%29%29%2F2
solutions:
x=-1%2F2%2Bsqrt%285%29%2F2
x=-1%2F2-sqrt%285%29%2F2