For the given polynomial approximate each zero to the nearest tenth. F (x)=x^3+4x^2-5x-20.

Question

X=?

(Type the interval or decimal rounded to the nearest tenth.Use a comma to separate answers as needed.)

Answer 1

Shantel,

 

There are a few ways to go about this.  I'll show one approach.

 

This is a cubic equation.  It may have 3 real roots or 1 real root and two complex conjugate roots.

 

As x tends to infinity, f(x) tends to infinity

As x tends to -infinity, f(x) tends to -infinity

For all x>=3, f(x)>0

f(2)=-6 so there must be at least one root in the interval (2,3)

 

A quadratic approximation for f(x+dx)=f(x)+dx.f'(x) +0.5f''(x).(dx)^2

 

f'(x)=3x^2+8x-5  f'(2)=23

f''(x)=6x+8          f''(2)=20

 

 

f(x+dx)=-6+23dx+10dx^2  

 

Notice also that f'>0 on the interval (2,3) so there can only be one root.  Let's find it:

 

If we use dx=0.1k where k is an integer, then if the sign of f changes between successive values of k, we shall have found a root within the required accuracy of 0.1...and I would prefer to express this as an interval as suggested by the question.

 

Trying -6+23*.1*k+10*.01k^2=-6+2.3k+.1k^2=0 , we find k is likely in the interval (2,3) and the root in (2.2,2.3)

 

Checking f(2.3)=1.827

              f(2.2)=-.992  and there is a sign change as required.

 

Since f'(x) has roots at  -4/3+-(1/6)sqrt(64+4*3*5)=-10/3 or 2/3, we know there is one root <-10/3 and one in the

interval(-10/3,2/3).  Let's try x=-4 and we find f(x)=0 so x=-4 is a root.  That was a piece of luck but we could have followed the same process as in the discovery of the first root.

 

Now that we have found one root exactly we can just factorize:  x^3+4x^2-5x-20=(x+4)(x^2-5)

So the other two roots are +/-sqrt5 and the unaccounted for root, -sqrt5 lies in the interval (-2.2,-2.3)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer 2

x³ + 4x² - 5x - 20 can be factored by grouping the first two terms and the last two:

 

x³ + 4x² - 5x - 20 =

(x³ + 4x²) - (5x + 20) =

x² (x + 4) - 5 (x + 4) =

(x² - 5) (x + 4)

 

Now we can see that the zeros are ±√5 and -4.

 

√5 to the nearest tenth is 2.2, so the zeros are -2.2, 2.2, and -4