Finding a polynomial function given zeros

[Illuvitar]

Hey guys I am having a little bit of trouble with using and understanding the linear factorization theorem to find the polynomial function.

Homework Statement

Find an nth degree polynomial function with real coefficents satisfying the given conditions.

n=3; -5 and 4+3i are zeros; f(2)=91

Homework Equations

The Attempt at a Solution

1) Since the polynomial has a degree of 3 I know there must be 3 linear factors which are:
(x+5) because -5 is a real zero and (x-4+3i) but also its conjugate (x-4-3i) are the complex zeros.

2) Now I multiply the 3 linear factors
(x-4+3i)(x-4-3i)= x²-4x-3xi-4x+16+12i+3ix-12i-3i²
and when I combine like term I get:

x²-8x+19 which I multiply by the remaining linear factor (x+5):

x³-3x²-21x+95

To find the function I apply f(2)=91

f(2)=a_n(2³-3(2)²-21(2)+95)=91 so

8-12-42+95=49a_n=91

I isolate the variable by dividing its coefficient and get:

a_n=91/49 simplified to 13/7

Now I substitute 13/7 for a_n and multiply by the product of the linear factors to get find the polynomial function:

f(x)=13/7(x³-3x²-21x+95) and end up with a messy looking product...anyway the real answer is:

f(x)=x³-3x²-15x+125

I have checked and double checked this problem several times to look for arithmetic mistakes but I just can't seem to match the right answer. I don't know if I just don't understand the concept (honestly I have a shaky understanding of zeros of polynomials in the first place, I am just learning about them today) or if I am just being sloppy with my arithmetic. Any help would be appreciated. Thank you.

 

[Ray Vickson]

Hey guys I am having a little bit of trouble with using and understanding the linear factorization theorem to find the polynomial function.

Homework Statement

Find an nth degree polynomial function with real coefficents satisfying the given conditions.

n=3; -5 and 4+3i are zeros; f(2)=91

Homework Equations

The Attempt at a Solution

1) Since the polynomial has a degree of 3 I know there must be 3 linear factors which are:
(x+5) because -5 is a real zero and (x-4+3i) but also its conjugate (x-4-3i) are the complex zeros.

2) Now I multiply the 3 linear factors
(x-4+3i)(x-4-3i)= x²-4x-3xi-4x+16+12i+3ix-12i-3i²
and when I combine like term I get:

x²-8x+19 which I multiply by the remaining linear factor (x+5):

x³-3x²-21x+95

To find the function I apply f(2)=91

f(2)=a_n(2³-3(2)²-21(2)+95)=91 so

8-12-42+95=49a_n=91

I isolate the variable by dividing its coefficient and get:

a_n=91/49 simplified to 13/7

Now I substitute 13/7 for a_n and multiply by the product of the linear factors to get find the polynomial function:

f(x)=13/7(x³-3x²-21x+95) and end up with a messy looking product...anyway the real answer is:

f(x)=x³-3x²-15x+125

I have checked and double checked this problem several times to look for arithmetic mistakes but I just can't seem to match the right answer. I don't know if I just don't understand the concept (honestly I have a shaky understanding of zeros of polynomials in the first place, I am just learning about them today) or if I am just being sloppy with my arithmetic. Any help would be appreciated. Thank you.

You should have $(x-4+3i)(x-4-3i) = x^2 -8x + 4^2 +3^2 = x^2 - 8x + 25$, not the $x^2 - 8x + 19$ that you claim.

 

[Illuvitar]

Oh crap. Thats so obvious. For some reason I thought 3i X-3i was -3i^2 not -9^2. Good lord I feel embarrassed. Thank for your help Ray.

 

[Steven60]

Also I feel that you were lucky that the complex roots come in conjugates pairs! If one root is 4+3i, the one factor must be [ x - (4+3i)], not [x-4+3i] as you wrote. Also there is a much better way of multiplying [ x - (4+3i)] and [ x - (4-3i)]. Write [ (x-4) + 3i ] * [ (x-4) - 3i]. Now thinking of (x-4) as A and 3i as B we are multiply (A+B)*(A-B) which is A^2 -B^2 or (x-4)^2 - (3i)^2 which is x^2-8x+16 +9 = x^2-8x+25.