SOLUTION: Verify that -2 is a root of the equation {{{2x^3+x^2-10x-8=0}}} Find the other two roots, correct to 2 decimal places.

Question 263378: Verify that -2 is a root of the equation
Find the other two roots, correct to 2 decimal places.
Found 2 solutions by drk, richwmiller:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
we can use synthetic division to verify. This will generate another equation that we will factor.
-2 // . . . 2 . . . . 1 . . . . -10 . . . . . -8
. . . . . . . . . . . . -4 . . . . .6 . . . . . . .8
. . . . . . . 2 . . . . -3 . . . . -4 . . . // . 0
So, we have verified by getting a 0 remainder. We now have a new equation of the form

By quadratic, we get

and then

we have two answers and rounded, we get
x ~ 2.35
x ~ -0.85
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
(x+2)(2x^2-3 x-4) = 0
there are several ways of veryfying that -2 is a solution
You can plug -2 in for x and see if the equation comes out equal
You can factor which I did.
and here are the other two solutions
by factoring

Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .

Now multiply the first coefficient by the last term to get .

Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?

To find these two numbers, we need to list all of the factors of (the previous product).

Factors of :

1,2,4,8

-1,-2,-4,-8

Note: list the negative of each factor. This will allow us to find all possible combinations.

These factors pair up and multiply to .

1*(-8) = -8
2*(-4) = -8
(-1)*(8) = -8
(-2)*(4) = -8

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :

First Number Second Number Sum
1 -8 1+(-8)=-7
2 -4 2+(-4)=-2
-1 8 -1+8=7
-2 4 -2+4=2

From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.

===============================================================

Answer:

So doesn't factor at all (over the rational numbers).

So is prime.

and using the quadratic formula

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=41 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 2.35078105935821, -0.850781059358212. Here's your graph:

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