This is driving me nuts!! Can someone please factor this polynomial please? 2x^3-x^2-5x-2=0 ??

Question

They also say one zero is 2, so one factor is (x-2)...I think....

Answer 1

You can use division, note I left out the x terms, but just imagine they are there.

So:
x - 2 is written as 1 -2
2x³ - x² - 5x - 2 is written as 2 -1 -5 -2
2x² + 3x + 1 is written as 2 3 1

.............2 ..3. 1
...... ---------------
1 -2 ) 2 -1 -5 -2
........ 2 -4
........-------
............. 3 -5
............. 3 -6
............. ------
................. 1 -2
................. 1 -2
..................------
...................... 0

That means:
2x³ - x² - 5x - 2 = 0

factors to:
(x - 2) (2x² + 3x + 1) = 0

You should be able to easily factor the equation now:
(x - 2)(2x + 1)(x + 1) = 0

So your solutions are:
x = 2
x = -½
x = -1

Answer 2

I think you are supposed to notice the coefficients (2, 1, 5, 2) and see that when x is +1 or -1, they COULD add to zero if all the signs came out the right way round. So try +1 and -1, and we do get the polynomial to be zero for x = -1, hence (x + 1) is a factor.

Synthetic division gives us (x + 1) (2x^2 - 3x - 2).

Trying to factor the quadratic as (2x ? 1) (x ? 2) it is easy to see which of the question marks needs to be a plus, and which a minus.

So finally 2x^3 - x^2 - 5x - 2 = (x + 1) (2x + 1) (x - 2) and the solutions are x = -1, x = -1/2, and x = 2.

Answer 3

2X^3-X^2-5X-2=0 ,If you substitute 2 for the value of X the equation will be satisfied this implies that it is divisible by( X-2).
dividing (2X^3-X^2-5X-2)by ( X-2).gives 2X^2+3X+1.
factorizing 2X^2+3X+1.gives (2X+1)(X+1).

(2X^3-X^2-5X-2)=(2X+1)(X+1)(X-2)

Answer 4

(x+1)(x+1/2)(x-2)= 0

you have to divide de inicial polynom by (x-2) and solve de cuadratic equation.

Answer 5

2x^3-X^2-5x-2=o
2x^3-2x-x^2-x-2x-2=0
2x[x^2-1]-x[x+1]-2[x+1]
2x[x+1][x-1]-x[x+1]-2[x+1]=0
[x+1][2x(x-1)-x-2]=0
[x+1][2x2-2x-x-2]
[x+1][2x^2-3x-2]=0
[x+1][2x^2-4x+x-2]=0
[x+1][2x(x-2)+1(x-2)]
[x+1][x-2][2x+1]=0
x=-1,2,-1/2