SOLUTION: Problem: Find all the real zeros of the polynomial. Use the quadratic formula if necessary. P(x)= x^3 + 4x^2 + 3x -2 I tried all the possible zeros with synthetic division

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Question 78251This question is from textbook College Algebra
: Problem: Find all the real zeros of the polynomial. Use the quadratic formula if necessary.
P(x)= x^3 + 4x^2 + 3x -2

I tried all the possible zeros with synthetic division and none came up as a zero. I think I am going wrong with division somewhere. Thank you.
This question is from textbook College Algebra

Answer by rmromero(383) About Me  (Show Source):
You can put this solution on YOUR website!

Problem: Find all the real zeros of the polynomial. Use the quadratic formula if necessary.
P(x)= x^3 + 4x^2 + 3x -2

To get the possible zeros of the polynomials p/q can help us, where p= 2
it is the last term. q = 1, this is the coefficient of the first term
We will factor p and q
p = 1*2 and -1*-2
= 1, 2, -1, -2
q = 1*1 and -1*-1
= 1, -1

possible zeros p/q are 1, -1, 2, -2
Test using synthetic division which among them are the real zeros of the polynomial
1 | 1 4 3 -2
1 5 8
_______________________
1 5 8 6 --- > not a real zero

-1 | 1 4 3 -2
-1 -3 0
_____________________
1 3 0 -2 ==> Not a real zero

2 | 1 4 3 -2
2 12 30
_____________________
1 6 15 28 ====> not a real zero

-2 | 1 4 3 -2
-2 -4 2
_________________
1 2 -1 0 ===> a Real zero
the factor now is (x + 2) (x^2 + 2x - 1)
|
--> we can use quadratic formula to get the factor for this where a = 1, b = 2 and c = -2
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-2+%2B-+sqrt%28+2%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+

x+=+%28-2+%2B-+sqrt%28+4+%2B4+%29%29%2F%282%29+
x+=+%28-2+%2B-+sqrt%288%29%29%2F%282%29+
x+=+%28-2+%2B-+2+%28sqrt%282%29%29%29%2F%282%29+
x+=+-1+%2B-+1+%28sqrt%282%29%29
x+=+-1+%2B-+sqrt%282%29
There are 3 real zeros of the polynomial
x= -2,
x+=+-1+%2B+sqrt%282%29,
x+=+-1+-+sqrt%282%29



Rmromero