SOLUTION: f(x)=2x^3+5x^2-x-6 I need to find the actual zeros on my calculator. Then, show the correct factorization of the function. Thanks.

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Question 1011576: f(x)=2x^3+5x^2-x-6
I need to find the actual zeros on my calculator. Then, show the correct factorization of the function. Thanks.
Found 2 solutions by ikleyn, Boreal:
Answer by ikleyn(52814)   (Show Source): You can put this solution on YOUR website!
.
f(x)=2x^3+5x^2-x-6
I need to find the actual zeros on my calculator. Then, show the correct factorization of the function. Thanks.
---------------------------------------------------------------- 
 = .

It is easy to check that x = 1 is the root.

Further, it means that the left side polynomial has a factor (x-1).

Make the long division of  by x-1.

You will get the quadratic polynomial as the quotient.

Then find its roots.

And the last thing. Do you know that if ,  and  are the roots of a polynomial of the third degree, 
then this polynomial is the product , 
where a is its leading coefficient at  ?

That is all.



Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
Enter the equation under Y=
use window -5<=x<=5
-10<=y<=10
scl 1
see what looks like a clear root at 1
Check that 2+5-1-6=0
(x-1) is a factor.
The other root is near -2. Check f(-2):-16+20+2-6
There is a third root that is not an integer.
divide x-1 into 2x^3+5x^2-x-6;
synthetic division
1/2==5==-1==-6
2===7==6===0
so the second factor is (2x^2+7x+6)
This factors into (2x+3)(x+2); function's factors are (x-1)(2x+3)(x+2)
roots are -2 and -1.5.
One can change the window to look at x between -1 and -2.
Use 2nd Trace to get calc, and 2 for zero. Put the cursor below the x-axis near where the function climbs above. Left bound: Enter. Move it to the right side and hit Enter for right bound. When Guess appears on screen, hit enter, and -1.5 will appear.
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