SOLUTION: find the zeros of a function y= (x^2+2x-7)(x^3+4x^2-21x) I forgot how to do these, can someone please help?

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Question 982451: find the zeros of a function
y= (x^2+2x-7)(x^3+4x^2-21x)
I forgot how to do these, can someone please help?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
y=0?

(x^2+2x-7)=0?
Either factorize further or use formula for general solution. Not factorable, so use formula for general solution of quadratic equation.

x^3+4x^2-21x=0?
First factor into x(x^2+4x-21)=0 and then either factorize the quadratic factor or use formula for general solution. (Factoring will work here).

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!

To find zeroes of a polynomial

y = %28x%5E2%2B2x-7%29.%28x%5E3%2B4x%5E2-21x%29

you need to find zeroes of the first factor  %28x%5E2%2B2x-7%29  and the the zeroes of the second factor  %28x%5E3%2B4x%5E2-21x%29.

The first factor  %28x%5E2%2B2x-7%29  has the zeroes  x%5B1%5D = -1+%2B+2sqrt%282%29  and  x%5B2%5D = -1+-+2sqrt%282%29  (use the quadratic formula).

The second factor is  %28x%5E3%2B4x%5E2-21x%29 = x.%28x%5E2%2B4x-21%29.

It has the zeroes  x%5B3%5D = 0,  x%5B4%5D = 4,  and  x%5B5%5D = -7.

The first of these three zeroes is obvious  (x%5B3%5D = 0).

The two others of these three zeroes you can obtain using the quadratic formula or the Viete's theorem
(see the lesson  Solving quadratic equations without quadratic formula  in this site).

Answer.  The zeroes are  -1+%2B+2sqrt%282%29,  -1+-+2sqrt%282%29,  0,  4  and  -7.