Find the zeros?

Could someone please help me with this, I gave my answer to the teacher and did not get one right. I'm still trying to figure it out. if you could help, it would be great, thanks.





Find all the zeros of the function and write the polynomial has a product of linear factors.


1. f(x) = x^3 + 11x^2 + 39x + 29


2. g(x) = x^5 – 8x^4 + 28x^3 – 56x^2 + 64x – 32








Find a polynomial function with integer coefficients that has the given zeros


3. 0, 0, 4, 1 + i





Use the given zero to find all the zeros of the function.


4. x^3 + x^2 + 9x + 9 zero r = 3i





a) use the root or zero feature of a graphing utility to aproximate the zero's of the function accurate to three decimal places


b)determine one of the exact zeros (use synthetic division) and find the exact value of the remaining zeros.


5. f(x) = x^3 + 4x^2 + 14x + 20

Find the zeros?
I'll try to get you started...





PROBLEM 1:





Here you need to "guess" at a factor.that will make it zero. Take the factors of the last coefficient (29) divided by factors of the first coefficient (1 on x^3). Don't forget negative factors.


So your possible factors are {-29, -1, 1, 29}.





Start with trying 1... no good because everything is positive. I suggest -1.





f(-1) = (-1)^3 + 11(-1)² + 39(-1) + 29


f(-1) = -1 + 11 - 39 + 29


f(-1) = 40 - 40


f(-1) = 0





So you know (x + 1) will be a factor. Now do synthetic division:


......................... x² + 10x + 29


x + 1 ) x^3 + 11x² + 39x + 29


........... x^3 + x²


..................... 10x² + 39x


..................... 10x² + 10x


.................................. 29x + 29


.................................. 29x + 29





Now you have:


f(x) = (x + 1)(x² + 10x + 29)





From here you need to use the quadratic formula to solve...





x² + 10x + 29 --%26gt; a = 1, b = 10, c = 29





x = -5 ± 4i





So the zeroes of the function are:


-1, -5 + 4i, -5 - 4i





As a product of linear factors:


f(x) = (x + 1)(x + 5 - 4i)(x + 5 + 4i)





PROBLEM 2:





Same method with a little more trial and error.





PROBLEM 3:





If the zeros are 0, 0, 4, 1+i





Then you start with (x - r1)(x - r2)(x - r3)(x - r4)





That becomes:


(x)(x)(x - 4)(x - (1+i))





Or:


x²(x - 4)(x - (1+i))





Note: since you have a root of 1+i, you should have a root of 1-i... so multiply that in too.


x²(x - 4)(x - (1+i))(x - (1-i))





x²(x - 4)(x -1 - i)(x-1 + i)


x²(x - 4)((x-1)² - i²)


x²(x - 4)(x² - 2x - 2 + 1)


x²(x - 4)(x² - 2x - 1)





Finally multiply it all out:


(x^3 - 4x²)(x² - 2x - 1)


(x^5 - 2x^4 - x^3) - (4x^4 - 8x^3 - 4x²)


x^5 - 2x^4 - 4x^4 - x^3 + 8x^3 + 4x²


= x^5 - 6x^4 + 7x^3 + 4x²
Reply:i dont have time to do this all by hand (sorry- have an honors chem test tomorrow!) but an easy way to find zeros is to put the equation into your graphing calculator, under y= instead of f(x) or g(x)=, and then see where the line crosses the x-axis, or where y=0. these points will be your zeros.


and for #4, try plugging in 3i for x, and determining all your possible answers


hope that helped a bit!
Reply:By inspection of (1), f(-1) = 0





synthetically divide out (x+1) from the polynomial








-1| 1 11 39 29


_______-1_-10__-29________


1 10 29 0





so the result is





f(x) = (x+1)(x^2 + 10x + 29)


use quadratic equation to find remaining zeros


x = (-10 +/-sqrt(10^2 - 4*29))/2 = (-10 +/- sqrt(-16))/2


x = -5 + 4i, -5 - 4i, -1 !!