Question

Find the zeros of the function f(x) = x^3 + 6x^2 + 15x + 10

Answer 1

Hello?

Answer 2

Wow you like answers fast!!

Answer 3

hahahaha sorry

Answer 4

Here they are:
-1
-2.5 + i* 1.9364916731037085
-2.5 - i* 1.9364916731037085

Answer 5

I'd explain how I got them but I don't want to waste any more time LOL !!!!!!!

Answer 6

Observing the function, \(x^3 + 6x^2 + 15x\) has to be \(-10\) so as to make the whole expression zero. From observation, you can see how \(-1 + 6 - 15 = -10\), thus making \(-1\) a root of the expression.

Answer 7

Yes, right! First find a solution ........

Answer 8

is -1 the only solution?

Answer 9

Let you have three solutions for this : \(\alpha , \beta ,\gamma \) then :

\((x-\alpha ) (x- \beta)( x- \gamma) = 0 \)

We got \(\alpha = -1\)

Thus,

\((x+1)(x -\beta)(x - \gamma) = 0 \)

Clearly : \((x-\beta)(x-\gamma) \) can be written as : \(x^2 + ax + b\) where a and b are numbers.

Answer 10

Thanks you all!

Answer 11

Use synthetic division now, to find a and b :

Write down the coefficients of the original expression as :

1 6 15 10 | x = -1
|
-----------------------------------------
1

I have written down the first coefficient (1)

Now multiply the "1" by "-1" = -1

Write this "-1" below the number "6"

1 6 15 10 | x = -1
-1
-----------------------------------------
1

Add 6 and -1

1 6 15 10 | x = -1
-1
-----------------------------------------
1 5

Now, again, multiply the "5" with "-1" :

1 6 15 10 | x = -1
-1 -5
-----------------------------------------
1 5

Add 15 and -5

1 6 15 10 | x = -1
-1 -5
-----------------------------------------
1 5 10

Similarly, multiply 10 with -1

1 6 15 10 | x = -1
-1 -5 -10
-----------------------------------------
1 5 10

Add 10 and -10

1 6 15 10 | x = -1
-1 -5 -10
-----------------------------------------
1 5 10 0

As the last answer we got is 0, thus "x = -1" is one root of the given equation.

Now, note that those : \(\bf{ 1 \space , 5 \space , 10 \space }\) are the coefficients of the expression : \(x^2 + ax + b\)

Thus,

you get :

\(x^2 + ax + b = 1(x^2) + 5(x) + 10\)

Above, we had :

\((x+1)(x^2 + ax+ b) = 0 \)

Thus, it becomes :

\((x+1)(x^2 + 5x + 10) =0 \)

One solution is x + 1 = 0 or x = - 1

Second and third solution will be obtained from : \(x^2 + 5x + 10 =0 \)

\(x = \cfrac{-5 \pm \sqrt{25 - 40}}{2} = \cfrac{ -5 \pm \sqrt{15} i }{2} \)


Or : x = \(\cfrac{-1}{2} \left( -5 + i\sqrt{15} \right) \) or x = \(\cfrac{-1}{2} \left( -5 - i\sqrt{15} \right) \) or x = -1.

I hope it helps.