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How to solve a third degree polynomial?

+2 votes
a^3 - 2a^2 - 4a + 8 = 0

im stuck. i totally forgot how to do this
asked Jan 11, 2013 in ALGEBRA 1 by anonymous Apprentice

2 Answers

+2 votes

a^3 - 2a^2 - 4a + 8 = 0

Take out common factors.

a^2 ( a - 2 ) - 4 ( a - 2 )= 0

Take out common factors.

(a^2-4)(a-2) = 0

(a^2 -2^2)(a-2) = 0

Note : A^2 - B^2 =  (A+B)(A-B)

(a-2)(a+2)(a-2) = 0

Simplify

a-2 = 0  or a+2 = 0  or a-2 = 0

a = 2    or   a = - 2   or  a = 2

There fore
 a = 2 ,- 2 ,and 2

a^3 - 2a^2 - 4a + 8 = 0  factors is (a-2)(a+2)(a-2)

answered Jan 11, 2013 by richardson Scholar
0 votes

Identify Rational Zeros  

Usually it is not practical to test all possible zeros of a polynomial function using only synthetic substitution. The Rational Zero Theorem can be used for finding the some possible zeros to test.

a3 - 2a2 - 4a + 8 = 0

If p/q is a rational zero, then p is a factor of 8 and q is a factor of 1.

The possible values of p are   ± 1,   ± 2, and   ± 4.

The possible values for q are ± 1.

So, p/q = ± 1,   ± 2, ± 4.

Make a table for the synthetic division and test possible  zeros.

p/q 1 -2 -4 8
1 1 -1 -5 3
2 1 0 -4 0

Since f(2) = 0, a = 2 is a zero. 

(a - 2) is a root.

The depressed polynomial is a2 - 4 = 0.

a2 - 4 = 0

a2 = 4

a = √ 4 = ±2.

The function has two real zeros at a = 2 and a = -2.

Therefore the roots are (a - 2) (a + 2).

The roots of a3 - 2a2 - 4a + 8 = 0 are (a - 2), (a - 2) and (a + 2).

 

answered Jul 2, 2014 by joly Scholar

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